Find the derivative of several variables online. Solving the derivative for dummies: definition, how to find, examples of solutions

Consider a function of two variables:

Since the variables $x$ and $y$ are independent, we can introduce the concept of a partial derivative for such a function:

The partial derivative of the function $f$ at the point $M=\left(((x)_(0));((y)_(0)) \right)$ with respect to the variable $x$ is the limit

\[(((f)")_(x))=\underset(\Delta x\to 0)(\mathop(\lim ))\,\frac(f\left(((x)_(0) )+\Delta x;((y)_(0)) \right))(\Delta x)\]

Similarly, we can define the partial derivative with respect to the variable $y$ :

\[(((f)")_(y))=\underset(\Delta y\to 0)(\mathop(\lim ))\,\frac(f\left(((x)_(0) );((y)_(0))+\Delta y \right))(\Delta y)\]

In other words, to find the partial derivative of a function of several variables, you need to fix all the other variables except the desired one, and then find the ordinary derivative with respect to this desired variable.

From this follows the main technique for calculating such derivatives: simply consider that all variables except the given one are constant, and then differentiate the function as you would differentiate the "ordinary" one - with one variable. For example:

$\begin(align)& ((\left(((x)^(2))+10xy \right))_(x))^(\prime )=((\left(((x)^(2 )) \right))^(\prime ))_(x)+10y\cdot ((\left(x \right))^(\prime ))_(x)=2x+10y, \\& (( \left(((x)^(2))+10xy \right))_(y))^(\prime )=((\left(((x)^(2)) \right))^(\ prime ))_(y)+10x\cdot ((\left(y \right))^(\prime ))_(y)=0+10x=10x. \\\end(align)$

Obviously, partial derivatives with respect to different variables give different answers - this is normal. It is much more important to understand why, say, in the first case, we calmly removed $10y$ from under the sign of the derivative, and in the second case, we completely nullified the first term. All this is due to the fact that all letters, except for the variable by which differentiation is carried out, are considered constants: they can be taken out, "burned", etc.

What is a "partial derivative"?

Today we will talk about functions of several variables and their partial derivatives. First, what is a function of multiple variables? Until now, we have been accustomed to think of a function as $y\left(x \right)$ or $t\left(x \right)$, or any variable and a single function from it. Now we will have one function, and several variables. When $y$ and $x$ change, the value of the function will change. For example, if $x$ doubles, the value of the function will change, while if $x$ changes and $y$ does not change, the value of the function will change in the same way.

Of course, a function of several variables, just like a function of one variable, can be differentiated. However, since there are several variables, it is possible to differentiate according to different variables. In this case, specific rules arise that did not exist when differentiating one variable.

First of all, when we consider the derivative of a function of any variable, we must indicate which variable we consider the derivative of - this is called the partial derivative. For example, we have a function of two variables, and we can calculate it both in $x$ and in $y$ - two partial derivatives of each of the variables.

Secondly, as soon as we have fixed one of the variables and begin to calculate the partial derivative with respect to it, then all the others included in this function are considered constants. For example, in $z\left(xy \right)$, if we consider the partial derivative with respect to $x$, then wherever we encounter $y$, we consider it a constant and treat it exactly as a constant. In particular, when calculating the derivative of a product, we can take $y$ out of the bracket (we have a constant), and when calculating the derivative of the sum, if we get somewhere the derivative of an expression containing $y$ and not containing $x$, then the derivative of this expression will be equal to "zero" as the derivative of the constant.

At first glance, it may seem that I am talking about something complex, and many students get confused at first. However, there is nothing supernatural in partial derivatives, and now we will see this on the example of specific problems.

Problems with radicals and polynomials

Task #1

In order not to waste time in vain, from the very beginning we will start with serious examples.

Let me start with the following formula:

This is the standard table value that we know from the standard course.

In this case, the derivative $z$ is calculated as follows:

\[(((z)")_(x))=((\left(\sqrt(\frac(y)(x)) \right))^(\prime ))_(x)=\frac( 1)(2\sqrt(\frac(y)(x)))((\left(\frac(y)(x) \right))^(\prime ))_(x)\]

Let's do it again, since the root is not $x$, but some other expression, in this case $\frac(y)(x)$, then first we will use the standard table value, and then, since the root is not $x $ and another expression, we need to multiply our derivative by one more of this expression with respect to the same variable. Let's start with the following:

\[((\left(\frac(y)(x) \right))^(\prime ))_(x)=\frac(((((y)"))_(x))\cdot x-y \cdot ((((x)"))_(x)))(((x)^(2)))=\frac(0\cdot x-y\cdot 1)(((x)^(2)) )=-\frac(y)(((x)^(2)))\]

We return to our expression and write:

\[(((z)")_(x))=((\left(\sqrt(\frac(y)(x)) \right))^(\prime ))_(x)=\frac( 1)(2\sqrt(\frac(y)(x)))((\left(\frac(y)(x) \right))^(\prime ))_(x)=\frac(1) (2\sqrt(\frac(y)(x)))\cdot \left(-\frac(y)(((x)^(2))) \right)\]

Basically, that's all. However, it is wrong to leave it in this form: such a construction is inconvenient to use for further calculations, so let's transform it a bit:

\[\frac(1)(2\sqrt(\frac(y)(x)))\cdot \left(-\frac(y)(((x)^(2))) \right)=\frac (1)(2)\cdot \sqrt(\frac(x)(y))\cdot \frac(y)(((x)^(2)))=\]

\[=-\frac(1)(2)\cdot \sqrt(\frac(x)(y))\cdot \sqrt(\frac(((y)^(2)))(((x)^ (4))))=-\frac(1)(2)\sqrt(\frac(x\cdot ((y)^(2)))(y\cdot ((x)^(4)))) =-\frac(1)(2)\sqrt(\frac(y)(((x)^(3))))\]

Answer found. Now let's deal with $y$:

\[(((z)")_(y))=((\left(\sqrt(\frac(y)(x)) \right))^(\prime ))_(y)=\frac( 1)(2\sqrt(\frac(y)(x)))\cdot ((\left(\frac(y)(x) \right))^(\prime ))_(y)\]

Let's write separately:

\[((\left(\frac(y)(x) \right))^(\prime ))_(y)=\frac(((((y)"))_(y))\cdot x-y \cdot ((((x)"))_(y)))(((x)^(2)))=\frac(1\cdot x-y\cdot 0)(((x)^(2)) )=\frac(1)(x)\]

Now we write:

\[(((z)")_(y))=((\left(\sqrt(\frac(y)(x)) \right))^(\prime ))_(y)=\frac( 1)(2\sqrt(\frac(y)(x)))\cdot ((\left(\frac(y)(x) \right))^(\prime ))_(y)=\frac( 1)(2\sqrt(\frac(y)(x)))\cdot \frac(1)(x)=\]

\[=\frac(1)(2)\cdot \sqrt(\frac(x)(y))\cdot \sqrt(\frac(1)(((x)^(2))))=\frac (1)(2)\sqrt(\frac(x)(y\cdot ((x)^(2))))=\frac(1)(2\sqrt(xy))\]

Done.

Task #2

This example is both simpler and more complex than the previous one. More difficult, because there are more actions, but easier, because there is no root and, in addition, the function is symmetric with respect to $x$ and $y$, i.e. if we swap $x$ and $y$, the formula doesn't change. This remark will further simplify the calculation of the partial derivative, i.e. it is enough to calculate one of them, and in the second one just swap $x$ and $y$.

Let's get down to business:

\[(((z)")_(x))=((\left(\frac(xy)(((x)^(2))+((y)^(2))+1) \right ))^(\prime ))_(x)=\frac(((\left(xy \right))^(\prime ))_(x)\left(((x)^(2))+( (y)^(2))+1 \right)-xy((\left(((x)^(2))+((y)^(2))+1 \right))^(\prime ) )_(x))(((\left(((x)^(2))+((y)^(2))+1 \right))^(2)))\]

Let's count:

\[((\left(xy \right))^(\prime ))_(x)=y\cdot ((\left(x \right))^(\prime ))=y\cdot 1=y\ ]

However, many students do not understand such a record, so we write it like this:

\[((\left(xy \right))^(\prime ))_(x)=((\left(x \right))^(\prime ))_(x)\cdot y+x\cdot ((\left(y \right))^(\prime ))_(x)=1\cdot y+x\cdot 0=y\]

Thus, we are once again convinced of the universality of the partial derivative algorithm: no matter how we consider them, if all the rules are applied correctly, the answer will be the same.

Now let's deal with one more partial derivative from our big formula:

\[((\left(((x)^(2))+((y)^(2))+1 \right))^(\prime ))_(x)=((\left((( x)^(2)) \right))^(\prime ))_(x)+((\left(((y)^(2)) \right))^(\prime ))_(x) +(((1)")_(x))=2x+0+0\]

We substitute the resulting expressions into our formula and get:

\[\frac(((\left(xy \right))^(\prime ))_(x)\left(((x)^(2))+((y)^(2))+1 \ right)-xy((\left(((x)^(2))+((y)^(2))+1 \right))^(\prime ))_(x))(((\left (((x)^(2))+((y)^(2))+1 \right))^(2)))=\]

\[=\frac(y\cdot \left(((x)^(2))+((y)^(2))+1 \right)-xy\cdot 2x)(((\left((( x)^(2))+((y)^(2))+1 \right))^(2)))=\]

\[=\frac(y\left(((x)^(2))+((y)^(2))+1-2((x)^(2)) \right))(((\ left(((x)^(2))+((y)^(2))+1 \right))^(2)))=\frac(y\left(((y)^(2)) -((x)^(2))+1 \right))(((\left(((x)^(2))+((y)^(2))+1 \right))^(2 )))\]

$x$ counted. And to calculate $y$ from the same expression, let's not perform the same sequence of actions, but use the symmetry of our original expression - we simply replace all $y$ in our original expression with $x$ and vice versa:

\[(((z)")_(y))=\frac(x\left(((x)^(2))-((y)^(2))+1 \right))((( \left(((x)^(2))+((y)^(2))+1 \right))^(2)))\]

Due to the symmetry, we calculated this expression much faster.

Nuances of the solution

For partial derivatives, all the standard formulas that we use for ordinary ones work, namely, the derivative of the private. In this case, however, its own specific features arise: if we consider the partial derivative of $x$, then when we get it from $x$, then we consider it as a constant, and therefore its derivative will be equal to "zero".

As in the case of ordinary derivatives, the quotient (the same) can be calculated in several different ways. For example, the same construction that we just calculated can be rewritten as follows:

\[((\left(\frac(y)(x) \right))^(\prime ))_(x)=y\cdot ((\left(\frac(1)(x) \right)) ^(\prime ))_(x)=-y\frac(1)(((x)^(2)))\]

\[((\left(xy \right))^(\prime ))_(x)=y\cdot (((x)")_(x))=y\cdot 1=y\]

However, on the other hand, you can use the formula from the derivative sum. As we know, it is equal to the sum of derivatives. For example, let's write the following:

\[((\left(((x)^(2))+((y)^(2))+1 \right))^(\prime ))_(x)=2x+0+0=2x \]

Now, knowing all this, let's try to work with more serious expressions, since real partial derivatives are not limited to polynomials and roots alone: ​​there are trigonometry, and logarithms, and an exponential function. Now let's do this.

Problems with trigonometric functions and logarithms

Task #1

We write the following standard formulas:

\[((\left(\sqrt(x) \right))^(\prime ))_(x)=\frac(1)(2\sqrt(x))\]

\[((\left(\cos x \right))^(\prime ))_(x)=-\sin x\]

Armed with this knowledge, let's try to solve:

\[(((z)")_(x))=((\left(\sqrt(x)\cdot \cos \frac(x)(y) \right))^(\prime ))_(x )=((\left(\sqrt(x) \right))^(\prime ))_(x)\cdot \cos \frac(x)(y)+\sqrt(x)\cdot ((\left (\cos \frac(x)(y) \right))^(\prime ))_(x)=\]

Let's write one variable separately:

\[((\left(\cos \frac(x)(y) \right))^(\prime ))_(x)=-\sin \frac(x)(y)\cdot ((\left( \frac(x)(y) \right))^(\prime ))_(x)=-\frac(1)(y)\cdot \sin \frac(x)(y)\]

Back to our design:

\[=\frac(1)(2\sqrt(x))\cdot \cos \frac(x)(y)+\sqrt(x)\cdot \left(-\frac(1)(y)\cdot \sin \frac(x)(y) \right)=\frac(1)(2\sqrt(x))\cdot \cos \frac(x)(y)-\frac(\sqrt(x))( y)\cdot \sin \frac(x)(y)\]

We have found everything for $x$, now let's do the calculations for $y$:

\[(((z)")_(y))=((\left(\sqrt(x)\cdot \cos \frac(x)(y) \right))^(\prime ))_(y )=((\left(\sqrt(x) \right))^(\prime ))_(y)\cdot \cos \frac(x)(y)+\sqrt(x)\cdot ((\left (\cos \frac(x)(y) \right))^(\prime ))_(y)=\]

Again, consider one expression:

\[((\left(\cos \frac(x)(y) \right))^(\prime ))_(y)=-\sin \frac(x)(y)\cdot ((\left( \frac(x)(y) \right))^(\prime ))_(y)=-\sin \frac(x)(y)\cdot x\cdot \left(-\frac(1)(( (y)^(2))) \right)\]

We return to the original expression and continue the solution:

\[=0\cdot \cos \frac(x)(y)+\sqrt(x)\cdot \frac(x)(((y)^(2)))\sin \frac(x)(y) =\frac(x\sqrt(x))(((y)^(2)))\cdot \sin \frac(x)(y)\]

Done.

Task #2

Let's write the formula we need:

\[((\left(\ln x \right))^(\prime ))_(x)=\frac(1)(x)\]

Now let's count by $x$:

\[(((z)")_(x))=((\left(\ln \left(x+\ln y \right) \right))^(\prime ))_(x)=\frac( 1)(x+\ln y).((\left(x+\ln y \right))^(\prime ))_(x)=\]

\[=\frac(1)(x+\ln y)\cdot \left(1+0 \right)=\frac(1)(x+\ln y)\]

Found by $x$. Counting by $y$:

\[(((z)")_(y))=((\left(\ln \left(x+\ln y \right) \right))^(\prime ))_(y)=\frac( 1)(x+\ln y).((\left(x+\ln y \right))^(\prime ))_(y)=\]

\[=\frac(1)(x+\ln y)\left(0+\frac(1)(y) \right)=\frac(1)(y\left(x+\ln y \right))\ ]

Problem solved.

Nuances of the solution

So, no matter what function we take the partial derivative of, the rules remain the same, regardless of whether we are working with trigonometry, with roots or with logarithms.

The classical rules for working with standard derivatives remain unchanged, namely, the derivative of the sum and difference, the quotient and the complex function.

The last formula is most often found in solving problems with partial derivatives. We meet them almost everywhere. There has not yet been a single task that we have not come across it there. But no matter what formula we use, we still add one more requirement, namely, the feature of working with partial derivatives. As soon as we fix one variable, all the others are constants. In particular, if we consider the partial derivative of the expression $\cos \frac(x)(y)$ with respect to $y$, then it is $y$ that is the variable, and $x$ remains constant everywhere. The same works vice versa. It can be taken out of the sign of the derivative, and the derivative of the constant itself will be equal to "zero".

All this leads to the fact that partial derivatives of the same expression, but with respect to different variables, can look completely different. For example, consider the following expressions:

\[((\left(x+\ln y \right))^(\prime ))_(x)=1+0=1\]

\[((\left(x+\ln y \right))^(\prime ))_(y)=0+\frac(1)(y)=\frac(1)(y)\]

Problems with exponential functions and logarithms

Task #1

Let's start by writing the following formula:

\[((\left(((e)^(x)) \right))^(\prime ))_(x)=((e)^(x))\]

Knowing this fact, as well as the derivative of a complex function, let's try to calculate. I will now solve in two different ways. The first and most obvious is the derivative of the product:

\[(((z)")_(x))=((\left(((e)^(x))\cdot ((e)^(\frac(x)(y))) \right) )^(\prime ))_(x)=((\left(((e)^(x)) \right))^(\prime ))_(x)\cdot ((e)^(\frac (x)(y)))+((e)^(x))\cdot ((\left(((e)^(\frac(x)(y))) \right))^(\prime ) )_(x)=\]

\[=((e)^(x))\cdot ((e)^(\frac(x)(y)))+((e)^(x))\cdot ((e)^(\frac (x)(y)))\cdot ((\left(\frac(x)(y) \right))^(\prime ))_(x)=\]

Let's solve the following expression separately:

\[((\left(\frac(x)(y) \right))^(\prime ))_(x)=\frac(((((x)"))_(x))\cdot y-x .((((y)"))_(x)))(((y)^(2)))=\frac(1\cdot y-x\cdot 0)(((y)^(2))) =\frac(y)(((y)^(2)))=\frac(1)(y)\]

We return to our original design and continue the solution:

\[=((e)^(x))\cdot ((e)^(\frac(x)(y)))+((e)^(x))\cdot ((e)^(\frac (x)(y)))\cdot \frac(1)(y)=((e)^(x))\cdot ((e)^(\frac(x)(y)))\left(1 +\frac(1)(y)\right)\]

Everything, $x$ counted.

However, as I promised, now we will try to calculate the same partial derivative in a different way. To do this, note the following:

\[((e)^(x))\cdot ((e)^(\frac(x)(y)))=((e)^(x+\frac(x)(y)))\]

Let's write it like this:

\[((\left(((e)^(x))\cdot ((e)^(\frac(x)(y))) \right))^(\prime ))_(x)=( (\left(((e)^(x+\frac(x)(y))) \right))^(\prime ))_(x)=((e)^(x+\frac(x)(y )))\cdot ((\left(x+\frac(x)(y) \right))^(\prime ))_(x)=((e)^(x+\frac(x)(y)) )\cdot \left(1+\frac(1)(y) \right)\]

As a result, we got exactly the same answer, but the amount of calculations turned out to be smaller. To do this, it was enough to notice that when the product is multiplied, the exponents can be added.

Now let's count by $y$:

\[(((z)")_(y))=((\left(((e)^(x))\cdot ((e)^(\frac(x)(y))) \right) )^(\prime ))_(y)=((\left(((e)^(x)) \right))^(\prime ))_(y)\cdot ((e)^(\frac (x)(y)))+((e)^(x))\cdot ((\left(((e)^(\frac(x)(y))) \right))^(\prime ) )_(y)=\]

\[=0\cdot ((e)^(\frac(x)(y)))+((e)^(x))\cdot ((e)^(\frac(x)(y))) \cdot ((\left(\frac(x)(y) \right))^(\prime ))_(y)=\]

Let's solve one expression separately:

\[((\left(\frac(x)(y) \right))^(\prime ))_(y)=\frac(((((x)"))_(y))\cdot y-x \cdot ((((y)"))_(y)))(((y)^(2)))=\frac(0-x\cdot 1)(((y)^(2))) =-\frac(1)(((y)^(2)))=-\frac(x)(((y)^(2)))\]

Let's continue the solution of our original construction:

\[=((e)^(x))\cdot ((e)^(\frac(x)(y)))\cdot \left(-\frac(x)(((y)^(2) )) \right)=-\frac(x)(((y)^(2)))\cdot ((e)^(x))\cdot ((e)^(\frac(x)(y) ))\]

Of course, the same derivative could be calculated in the second way, the answer would be the same.

Task #2

Let's count by $x$:

\[(((z)")_(x))=((\left(x \right))_(x))\cdot \ln \left(((x)^(2))+y \right )+x\cdot ((\left(\ln \left(((x)^(2))+y \right) \right))^(\prime ))_(x)=\]

Let's count one expression separately:

\[((\left(\ln \left(((x)^(2))+y \right) \right))^(\prime ))_(x)=\frac(1)(((x )^(2))+y)\cdot ((\left(((x)^(2))+y \right))^(\prime ))_(x)=\frac(2x)((( x)^(2))+y)\]

Let's continue the solution of the original construction: $$

Here is the answer.

It remains to find by analogy by $y$:

\[(((z)")_(y))=((\left(x \right))^(\prime ))_(y).\ln \left(((x)^(2)) +y \right)+x\cdot ((\left(\ln \left(((x)^(2))+y \right) \right))^(\prime ))_(y)=\]

Let's count one expression separately as always:

\[((\left(((x)^(2))+y \right))^(\prime ))_(y)=((\left(((x)^(2)) \right) )^(\prime ))_(y)+(((y)")_(y))=0+1=1\]

We continue the solution of the main structure:

Everything is counted. As you can see, depending on which variable is taken for differentiation, the answers are completely different.

Nuances of the solution

Here is a vivid example of how the derivative of the same function can be calculated in two different ways. Look here:

\[(((z)")_(x))=\left(((e)^(x))\cdot ((e)^(\frac(x)(y))) \right)=( (\left(((e)^(x)) \right))^(\prime ))_(x)\cdot ((e)^(\frac(x)(y)))+((e) ^(x))\cdot ((\left(((e)^(\frac(x)(y))) \right))^(\prime ))_(x)=\]

\[=((e)^(x))\cdot ((e)^(\frac(x)(y)))+((e)^(x))\cdot ((e)^(\frac (x)(y)))\cdot \frac(1)(y)=((e)^(x))\cdot ((e)^(^(\frac(x)(y))))\ left(1+\frac(1)(y)\right)\]

\[(((z)")_(x))=((\left(((e)^(x)).((e)^(\frac(x)(y))) \right)) ^(\prime ))_(x)=((\left(((e)^(x+\frac(x)(y))) \right))^(\prime ))_(x)=(( e)^(x+\frac(x)(y))).((\left(x+\frac(x)(y) \right))^(\prime ))_(x)=\]

\[=((e)^(x))\cdot ((e)^(^(\frac(x)(y))))\left(1+\frac(1)(y) \right)\ ]

When choosing different paths, the amount of calculations may be different, but the answer, if everything is done correctly, will be the same. This applies to both classical and partial derivatives. At the same time, I remind you once again: depending on which variable the derivative is taken from, i.e. differentiation, the answer can be completely different. Look:

\[((\left(\ln \left(((x)^(2))+y \right) \right))^(\prime ))_(x)=\frac(1)(((x )^(2))+y)\cdot ((\left(((x)^(2))+y \right))^(\prime ))_(x)=\frac(1)((( x)^(2))+y)\cdot 2x\]

\[((\left(\ln \left(((x)^(2))+y \right) \right))^(\prime ))_(y)=\frac(1)(((x )^(2))+y)\cdot ((\left(((x)^(2))+y \right))^(\prime ))_(y)=\frac(1)((( x)^(2))+y)\cdot 1\]

In conclusion, to consolidate all this material, let's try to count two more examples.

Problems with a trigonometric function and a function with three variables

Task #1

Let's write these formulas:

\[((\left(((a)^(x)) \right))^(\prime ))=((a)^(x))\cdot \ln a\]

\[((\left(((e)^(x)) \right))^(\prime ))=((e)^(x))\]

Let's now solve our expression:

\[(((z)")_(x))=((\left(((3)^(x\sin y)) \right))^(\prime ))_(x)=((3 )^(x.\sin y))\cdot \ln 3\cdot ((\left(x\cdot \sin y \right))^(\prime ))_(x)=\]

Separately, consider the following construction:

\[((\left(x\cdot \sin y \right))^(\prime ))_(x)=(((x)")_(x))\cdot \sin y+x((\ left(\sin y \right))^(\prime ))_(x)=1\cdot \sin y+x\cdot 0=\sin y\]

We continue to solve the original expression:

\[=((3)^(x\sin y))\cdot \ln 3\cdot \sin y\]

This is the final private variable answer for $x$. Now let's count by $y$:

\[(((z)")_(y))=((\left(((3)^(x\sin y)) \right))^(\prime ))_(y)=((3 )^(x\sin y))\cdot \ln 3\cdot ((\left(x\sin y \right))^(\prime ))_(y)=\]

Let's solve one expression separately:

\[((\left(x\cdot \sin y \right))^(\prime ))_(y)=(((x)")_(y))\cdot \sin y+x((\ left(\sin y \right))^(\prime ))_(y)=0\cdot \sin y+x\cdot \cos y=x\cdot \cos y\]

We solve our construction to the end:

\[=((3)^(x\cdot \sin y))\cdot \ln 3\cdot x\cos y\]

Task #2

At first glance, this example may seem rather complicated, because there are three variables. In fact, this is one of the easiest tasks in today's video tutorial.

Find by $x$:

\[(((t)")_(x))=((\left(x((e)^(y))+y((e)^(z)) \right))^(\prime ) )_(x)=((\left(x\cdot ((e)^(y)) \right))^(\prime ))_(x)+((\left(y\cdot ((e) ^(z)) \right))^(\prime ))_(x)=\]

\[=((\left(x \right))^(\prime ))_(x)\cdot ((e)^(y))+x\cdot ((\left(((e)^(y )) \right))^(\prime ))_(x)=1\cdot ((e)^(y))+x\cdot o=((e)^(y))\]

Now let's deal with $y$:

\[(((t)")_(y))=((\left(x\cdot ((e)^(y))+y\cdot ((e)^(z)) \right))^ (\prime ))_(y)=((\left(x\cdot ((e)^(y)) \right))^(\prime ))_(y)+((\left(y\cdot ((e)^(z)) \right))^(\prime ))_(y)=\]

\[=x\cdot ((\left(((e)^(y)) \right))^(\prime ))_(y)+((e)^(z))\cdot ((\left (y \right))^(\prime ))_(y)=x\cdot ((e)^(y))+((e)^(z))\]

We have found the answer.

Now it remains to find by $z$:

\[(((t)")_(z))=((\left(x\cdot ((e)^(y))+((y)^(z)) \right))^(\prime ))_(z)=((\left(x\cdot ((e)^(y)) \right))^(\prime ))_(z)+((\left(y\cdot ((e )^(z)) \right))^(\prime ))_(z)=0+y\cdot ((\left(((e)^(z)) \right))^(\prime )) _(z)=y\cdot ((e)^(z))\]

We have calculated the third derivative, on which the solution of the second problem is completely completed.

Nuances of the solution

As you can see, there is nothing complicated in these two examples. The only thing we have seen is that the derivative of a complex function is often used, and depending on which partial derivative we consider, we get different answers.

In the last task, we were asked to deal with a function of three variables at once. There is nothing wrong with this, but at the very end we made sure that they all differ significantly from each other.

Key points

The final conclusions from today's video tutorial are as follows:

1. Partial derivatives are considered in the same way as ordinary ones, while in order to calculate the partial derivative with respect to one variable, all other variables included in this function, we take as constants.
2. When working with partial derivatives, we use all the same standard formulas as with ordinary derivatives: the sum, the difference, the derivative of the product and the quotient, and, of course, the derivative of a complex function.

Of course, watching this video tutorial alone is not enough to fully understand this topic, so right now on my website for this particular video there is a set of tasks dedicated to today's topic - go, download, solve these tasks and check the answer. And after that, you will not have any problems with partial derivatives either in exams or in independent work. Of course, this is far from the last lesson in higher mathematics, so visit our website, add VKontakte, subscribe to YouTube, put likes and stay with us!

The concept of a function of many variables

Let there be n-variables and each x 1, x 2 ... x n from a certain set x is assigned a definition. the number Z, then on the set x the function Z \u003d f (x 1, x 2 ... x n) of many variables is given.

X - area of ​​\u200b\u200bdefined functions

x 1, x 2 ... x n - independent variable (arguments)

Z - function Example: Z \u003d P x 2 1 * x 2 (Cylinder volume)

Consider Z \u003d f (x; y) - f-tion of 2 variables x (x 1, x 2 replaced by x, y). The results are by analogy transferred to other functions of many variables. The area of ​​\u200b\u200bdefining the function of 2 variables is the entire cord of the square (ooh) or part of it. Mn-in the value of the th function of 2 variables - the surface in a 3-dimensional space.

Techniques for constructing graphs: - Rassm-t section over the surface of the square || coordinate squares.

Example: x \u003d x 0, zn. square X || 0yz y \u003d y 0 0xz Type of function: Z \u003d f (x 0, y); Z=f(x, y 0)

For example: Z=x 2 +y 2 -2y

Z= x 2 +(y-1) 2 -1 x=0 Z=(y-1) 2 -1 y=1 Z= x 2 -1 Z=0 x 2 +(y-1) 2 -1

Parabola circle(center(0;1)

Limits and continuity of functions of two variables

Let Z = f (x; y) be given, then A is the limit of the f-tion in m. (x 0, y 0), if for any arbitrarily small put. number E>0 noun-t positive number b>0, that for all x,y satisfying |x-x 0 |<б; |y-y 0 |<б выполняется нерав-во |f(x,y)-A|

Z \u003d f (x; y) is continuous in t. (x 0, y 0), if: - it is defined in this t .; - has a finite limit at x, tending to x 0 and y to y 0; - this limit = value

functions in t. (x 0, y 0), i.e. limf (x; y) \u003d f (x 0, y 0)

If the function is continuous in each. t. mn-va X, then it is continuous in this area

Differential function, its geomeaning. The use of dif-la in approximate values.

dy=f’(x)∆x – differential function

dy=dx, i.e. dy=f '(x)dx if y=x

From a geologist's point of view, a function differential is an increment in the ordinate of the tangent drawn to the graph of the function at a point with the abscissa x 0

Dif-l is used in the calculation of approx. function values ​​according to the formula: f(x 0 +∆x)~f(x 0)+f’(x 0)∆x

The closer ∆x is to x, the more accurate the result.

Partial derivatives of the first and second order

First order derivative (which is called private)

A. Let x, y be the increments of independent variables x and y at some point from the region X. Then the value equal to z = f(x + x, y + y) = f(x, y) is called the total increment at the point x 0, y 0. If the variable x is fixed, and the variable y is incremented by y, then we get zу = f(x, y, + y) – f(x, y)

The partial derivative of the variable y is defined similarly, i.e.

The partial derivative of a function of 2 variables is found according to the same rules as for functions of one variable.

The difference is that when differentiating a function with respect to the variable x, y is considered const, and when differentiating with respect to y, x is considered const.

Isolated consts are connected to the function with addition/subtraction operations.

The associated consts are connected to the function with multiplication/division operations.

Derivative of isolated const = 0

1.4.Total differential of a function of 2 variables and its applications

Let z = f(x,y), then

tz = - is called a full increment

Partial derivative of the 2nd order

For continuous functions of 2 variables, the mixed partial derivatives of the 2nd order and coincide.

The use of partial derivatives to determine the partial derivatives of max and min functions are called extrema.

A. Points are called max or min z = f(x,y) if there are some segments such that for all x and y from this neighborhood f(x,y)

T. If an extremum point of a function of 2 variables is given, then the value of partial derivatives at this point is equal to 0, i.e. ,

The points at which first-order partial derivatives are called stationary or critical.

Therefore, to find the extremum points of a function of 2 variables, sufficient extremum conditions are used.

Let the function z = f(x,y) be twice differentiable, and let the stationary point,

1) , and maxA<0, minA>0.

1.4.(*)full differential. The geometric meaning of the differential. Application of the differential in approximate calculations

O. Let the function y = f(x) be defined in some neighborhood at the points . A function f(x) is called differentiable at a point if its increment at this point , where is represented in the form (1)

Where A is a constant value independent of , at a fixed point x, - infinitely small at . A relatively linear function A is called the differential of the function f(x) at a point and is denoted by df() or dy.

Thus, expression (1) can be written as ().

The function differential in expression (1) has the form dy = A . Like any linear function, it is defined for any value while the increment of the function must be considered only for those for which + belongs to the domain of the function f(x).

For convenience of notation of the differential, the increment is denoted by dx and is called the differential of the independent variable x. Therefore, the differential is written as dy = Adx.

If the function f(x) is differentiable at every point of some interval, then its differential is a function of two variables - the point x and the variable dx:

T. In order for the function y = g(x) to be differentiable at some point , it is necessary and sufficient that it has a derivative at this point, while

(*)Proof. Necessity.

Let the function f(x) be differentiable at the point , i.e., . Then

Therefore, the derivative f'() exists and is equal to A. Hence dy = f'()dx

Adequacy.

Let there be a derivative f'(), i.e. = f'(). Then the curve y = f(x) is a tangent segment. To calculate the value of a function at a point x, take a point in some of its neighborhood, such that it is not difficult to find f() and f’()/

Definition 1.11 Let a function of two variables be given z=z(x,y), (x,y)D . Dot M 0 (x 0 ;y 0 ) - internal point of the area D .

If in D there is such a neighborhood UM 0 points M 0 , which for all points

then point M 0 is called the local maximum point. But the meaning itself z(M 0 ) - local maximum.

But if for all points

then point M 0 is called the local minimum point of the function z(x,y) . But the meaning itself z(M 0 ) - local minimum.

The local maximum and local minimum are called local extrema of the function z(x,y) . On fig. 1.4 explains the geometric meaning of the local maximum: M 0 is the maximum point, since on the surface z=z(x,y) its corresponding point C 0 is above any neighboring point C (this is the locality of the maximum).

Note that there are points on the surface as a whole (for example, IN ) that are above C 0 , but these points (for example, IN ) are not "adjacent" to point C 0 .

In particular, the point IN corresponds to the concept of a global maximum:

The global minimum is defined similarly:

Finding global maxima and minima will be discussed in Section 1.10.

Theorem 1.3(necessary extremum conditions).

Let the function z =z(x,y),(x,y)D . Dot M 0 (x 0 ;y 0 D - local extremum point.

If at this point there are z" x And z" y , That

The geometric proof is "obvious". If at the point C 0 on (Fig. 1.4) to draw a tangent plane, then it "naturally" will pass horizontally, i.e. at an angle 0° to the axis Oh and to the axis OU .

Then, in accordance with the geometric meaning of partial derivatives (Fig. 1.3):

which was to be proved.

Definition 1.12.

If at the point M 0 conditions (1.41) are satisfied, then it is called a stationary point of the function z (x,y) .

Theorem 1.4(sufficient conditions for an extremum).

Let z =z(x,y),(x,y)D , which has second-order partial derivatives in some neighborhood of the point M 0 (x 0 ,y 0 )D . And M 0 - stationary point (i.e., the necessary conditions (1.41) are satisfied). Let's calculate:

The proof of the theorem uses topics (Taylor's formula for functions of several variables and the theory of quadratic forms) that are not covered in this tutorial.

Example 1.13.

Explore to the extreme:

Solution

1. Find stationary points by solving system (1.41):

that is, four stationary points are found. 2.

by Theorem 1.4 at a point is a minimum. And

by Theorem 1.4 at the point

Maximum. And

The general principle of finding second-order partial derivatives of a function of three variables is similar to the principle of finding second-order partial derivatives of a function of two variables.

In order to find the partial derivatives of the second order, you must first find the partial derivatives of the first order or, in another notation:

There are nine partial derivatives of the second order.

The first group is the second derivatives with respect to the same variables:

Or - the second derivative with respect to "x";

Or - the second derivative with respect to "y";

Or - the second derivative with respect to "z".

The second group is mixed partial derivatives of the 2nd order, there are six of them:

Or - mixed derivative "by x y";

Or - mixed derivative "by y x";

Or - mixed derivative "by x z";

Or - mixed derivative "po zet x";

Or - mixed derivative "by game z";

Or - mixed derivative "po z y".

As in the case of a function of two variables, when solving problems, one can focus on the following equalities of mixed second-order derivatives:

Note: Strictly speaking, this is not always the case. For the equality of mixed derivatives, it is necessary to fulfill the requirement of their continuity.

Just in case, a few examples of how to read this disgrace out loud:

- "two strokes twice a y";

- “de two y po de zet square”;

- “two strokes on x on z”;

- “de two y po de z po de y”.

Example 10

Find all first and second order partial derivatives for a function of three variables:

.

Solution: First, we find the partial derivatives of the first order:

We take the found derivative

and differentiate it by "y":

We take the found derivative

and differentiate it by "x":

Equality is done. Fine.

We deal with the second pair of mixed derivatives.

We take the found derivative

and differentiate it by "z":

We take the found derivative

and differentiate it by "x":

Equality is done. Fine.

Similarly, we deal with the third pair of mixed derivatives:

Equality is done. Fine.

After the work done, it is guaranteed that, firstly, we correctly found all partial derivatives of the 1st order, and secondly, we also correctly found the mixed partial derivatives of the 2nd order.

It remains to find three more partial derivatives of the second order, here, in order to avoid errors, you should concentrate as much as possible:

Ready. Again, the task is not so much difficult as voluminous. The solution can be shortened and referred to as equalities of mixed partial derivatives, but in this case there will be no verification. So it's better to take the time and find All derivatives (besides, this may be required by the teacher), or, in extreme cases, check on a draft.

Example 11

Find all first and second order partial derivatives for a function of three variables

.

This is a do-it-yourself example.

Solutions and answers:

Example 2:Solution:

Example 4:Solution: Let us find partial derivatives of the first order.

We compose the total differential of the first order:

Example 6:Solution: M(1, -1, 0):

Example 7:Solution: Let us calculate the partial derivatives of the first order at the pointM(1, 1, 1):

Example 9:Solution:

Example 11:Solution: Let's find partial derivatives of the first order:

Let's find partial derivatives of the second order:

.

Integrals

8.1. Indefinite integral. Detailed Solution Examples

Let's start studying the topic Indefinite integral", and also analyze in detail examples of solutions to the simplest (and not quite) integrals. As usual, we will limit ourselves to the minimum theory that is in numerous textbooks, our task is to learn how to solve integrals.

What do you need to know to successfully master the material? In order to cope with integral calculus, you need to be able to find derivatives, at least at an average level. It will not be superfluous experience if you have several dozen, or better, a hundred independently found derivatives behind you. At the very least, you should not be confused by the task of differentiating the simplest and most common functions.

It would seem, where are the derivatives at all, if we are talking about integrals in the article?! And here's the thing. The fact is that finding derivatives and finding indefinite integrals (differentiation and integration) are two mutually inverse actions, such as addition / subtraction or multiplication / division. Thus, without a skill and some kind of experience in finding derivatives, unfortunately, one cannot advance further.

In this regard, we will need the following methodological materials: Derivative table And Table of integrals.

What is the difficulty of studying indefinite integrals? If in derivatives there are strictly 5 rules of differentiation, a table of derivatives and a fairly clear algorithm of actions, then in integrals everything is different. There are dozens of integration methods and techniques. And, if the integration method was initially chosen incorrectly (that is, you don’t know how to solve it), then the integral can be “pricked” literally for days, like a real rebus, trying to notice various tricks and tricks. Some even like it.

By the way, we quite often heard from students (not humanities) an opinion like: “I have never had an interest in solving the limit or derivative, but integrals are a completely different matter, it’s exciting, there is always a desire to “break” a complex integral” . Stop. Enough black humor, let's move on to these very indefinite integrals.

Since there are many ways to solve, then where does a teapot start studying indefinite integrals? In integral calculus, in our opinion, there are three pillars or a kind of "axis" around which everything else revolves. First of all, you should have a good understanding of the simplest integrals (this article).

Then you need to work out the lesson in detail. THIS IS THE MOST IMPORTANT RECEPTION! Perhaps even the most important article of all articles devoted to integrals. And thirdly, be sure to read integration by parts, because it integrates a wide class of functions. If you master at least these three lessons, then there are already “not two”. You can be forgiven for not knowing integrals of trigonometric functions, integrals of fractions, integrals of fractional rational functions, integrals of irrational functions (roots), but if you “get into a puddle” on the replacement method or the integration by parts method, then it will be very, very bad.

So, let's start simple. Let's look at the table of integrals. As in derivatives, we notice several integration rules and a table of integrals of some elementary functions. Any tabular integral (and indeed any indefinite integral) has the form:

Let's get straight to the notation and terms:

- integral icon.

- integrand function (written with the letter "s").

– differential icon. What it is, we will consider very soon. The main thing is that when writing the integral and during the solution, it is important not to lose this icon. There will be a noticeable flaw.

is the integrand or "stuffing" of the integral.

– antiderivative function.

– . There is no need to be heavily loaded with terms, the most important thing here is that in any indefinite integral, a constant is added to the answer.

To solve an indefinite integral means to findset of antiderivative functions from the given integrand

Let's take a look at the entry again:

Let's look at the table of integrals.

What's happening? Our left parts are turning to other functions: .

Let's simplify our definition:

Solve the indefinite integral - it means to TURN it into an indefinite (up to a constant) function , using some rules, techniques and a table.

Take, for example, the table integral . What happened? The symbolic record has turned into a set of antiderivative functions.

As in the case of derivatives, in order to learn how to find integrals, it is not necessary to be aware of what an integral is, or an antiderivative function from a theoretical point of view. It is enough just to carry out transformations according to some formal rules. So, in case it is not at all necessary to understand why the integral turns into exactly. You can take this and other formulas for granted. Everyone uses electricity, but few people think about how electrons run along the wires.

Since differentiation and integration are opposite operations, for any antiderivative that is found correctly, the following is true:

In other words, if the correct answer is differentiated, then the original integrand must be obtained.

Let's go back to the same table integral .

Let's verify the validity of this formula. We take the derivative of the right side:

is the original integrand.

By the way, it became clearer why a constant is always assigned to a function. When differentiating, a constant always turns into zero.

Solve the indefinite integral it means to find a bunch of all antiderivatives, and not some single function. In the considered tabular example, , , , etc. - all these functions are the solution of the integral . There are infinitely many solutions, so they write briefly:

Thus, any indefinite integral is easy enough to check. This is some compensation for a large number of integrals of different types.

Let's move on to specific examples. Let's start, as in the study of the derivative, with two rules of integration:

- constant C can (and should) be taken out of the integral sign.

– the integral of the sum (difference) of two functions is equal to the sum (difference) of two integrals. This rule is valid for any number of terms.

As you can see, the rules are basically the same as for derivatives. Sometimes they are called linearity properties integral.

Example 1

Find the indefinite integral.

Run a check.

Solution: It is more convenient to convert it like.

(1) Applying the rule . Don't forget to write down the differential icon dx under each integral. Why under each? dxis a full multiplier. If you paint in detail, then the first step should be written as follows:

.

(2) According to the rule we take all the constants out of the signs of the integrals. Note that in the last term tg 5 is a constant, we also take it out.

In addition, at this step we prepare the roots and degrees for integration. In the same way as in differentiation, the roots must be represented in the form . Roots and degrees that are located in the denominator - move up.

Note: unlike derivatives, roots in integrals do not always need to be reduced to the form , and move the degrees up.

For example, - this is a ready-made tabular integral, which has already been calculated before you, and all sorts of Chinese tricks like completely unnecessary. Similarly: - this is also a tabular integral, there is no point in representing a fraction in the form . Study the table carefully!

(3) All integrals are tabular. We carry out the transformation using the table, using the formulas: , And

for a power function - .

It should be noted that the table integral is a special case of the formula for a power function: .

Constant C just add it once at the end of the expression

(rather than putting them after each integral).

(4) We write the result obtained in a more compact form, when all degrees of the form

again represent as roots, and the powers with a negative exponent are reset back to the denominator.

Examination. In order to perform the check, you need to differentiate the received answer:

Initial integrand, i.e., the integral was found correctly. From what they danced, to that they returned. It's good when the story with the integral ends just like that.

From time to time, there is a slightly different approach to checking the indefinite integral, when not the derivative, but the differential is taken from the answer:

.

As a result, we obtain not an integrand, but an integrand.

Do not be afraid of the concept of differential.

The differential is the derivative multiplied by dx.

However, it is not theoretical subtleties that are important to us, but what to do next with this differential. The differential is revealed as follows: icon d remove, put a stroke on the right above the bracket, assign a multiplier at the end of the expression dx :

Received initial integrand, that is, the integral is found correctly.

As you can see, the differential comes down to finding the derivative. I like the second way of checking less, since I have to additionally draw large brackets and drag the differential icon dx until the end of the test. Although it is more correct, or "more solid", or something.

In fact, it was possible to keep silent about the second method of verification. The point is not in the method, but in the fact that we have learned to open the differential. Again.

The differential is revealed as follows:

1) icon d remove;

2) put a stroke on the right above the bracket (the designation of the derivative);

3) at the end of the expression we assign a factor dx .

For example:

Remember this. We will need the considered technique very soon.

Example 2

.

When we find an indefinite integral, we ALWAYS try to check Moreover, there is a great opportunity for this. Not all types of problems in higher mathematics are a gift from this point of view. It does not matter that verification is often not required in control tasks, no one, and nothing prevents it from being carried out on a draft. An exception can be made only when there is not enough time (for example, at the test, exam). Personally, I always check integrals, and I consider the lack of verification to be a hack and a poorly completed task.

Example 3

Find the indefinite integral:

. Run a check.

Solution: Analyzing the integral, we see that under the integral we have the product of two functions, and even the exponentiation of the whole expression. Unfortunately, in the field of integral battle No good and comfortable formulas for integrating the product and the quotient as: or .

Therefore, when a product or a quotient is given, it always makes sense to see if it is possible to transform the integrand into a sum? The considered example is the case when it is possible.

First, we give the complete solution, the comments will be below.

(1) We use the good old formula for the square of the sum for any real numbers, getting rid of the degree above the common bracket. outside the brackets and applying the abbreviated multiplication formula in the opposite direction: .

Example 4

Find the indefinite integral

Run a check.

This is an example for self-solving. Answer and complete solution at the end of the lesson.

Example 5

Find the indefinite integral

. Run a check.

In this example, the integrand is a fraction. When we see a fraction in the integrand, the first thought should be the question: “Is it possible to somehow get rid of this fraction, or at least simplify it?”.

We notice that the denominator contains a lone root of "x". One in the field is not a warrior, which means that you can divide the numerator into the denominator term by term:

We do not comment on actions with fractional powers, since they have been repeatedly discussed in articles on the derivative of a function.

If you are still confused by such an example as

and no one gets the right answer,

Also note that the solution skips one step, namely applying the rules , . Usually, with a certain experience in solving integrals, these rules are considered an obvious fact and are not described in detail.

Example 6

Find the indefinite integral. Run a check.

This is an example for self-solving. Answer and complete solution at the end of the lesson.

In the general case, with fractions in integrals, everything is not so simple, additional material on the integration of fractions of some types can be found in the article: Integration of some fractions. But, before moving on to the above article, you need to read the lesson: Replacement method in indefinite integral. The fact is that summing a function under a differential or a variable change method is key point in the study of the topic, since it is found not only "in pure assignments for the replacement method", but also in many other varieties of integrals.

Solutions and answers:

Example 2: Solution:

Example 4: Solution:

In this example, we used the reduced multiplication formula

Example 6: Solution:

The method of changing a variable in an indefinite integral. Solution examples

In this lesson, we will get acquainted with one of the most important and most common tricks that is used in the course of solving indefinite integrals - the change of variable method. For successful mastering of the material, initial knowledge and integration skills are required. If there is a feeling of an empty full teapot in integral calculus, then you should first familiarize yourself with the material Indefinite integral. Solution examples, where it is explained in an accessible form what an integral is and basic examples for beginners are analyzed in detail.

Technically, the method of changing a variable in an indefinite integral is implemented in two ways:

– Bringing the function under the sign of the differential.

– The actual change of variable.

In fact, it's the same thing, but the design of the solution looks different. Let's start with a simpler case.

We continue the favorite topic of mathematical analysis - derivatives. In this article, we will learn how to find partial derivatives of a function of three variables: first derivatives and second derivatives. What do you need to know and be able to master the material? Do not believe it, but, firstly, you need to be able to find the "ordinary" derivatives of a function of one variable - at a high or at least an average level. If it’s really tight with them, then start with a lesson How to find the derivative? Secondly, it is very important to read the article and comprehend and solve, if not all, then most of the examples. If this has already been done, then walk with me with a confident gait, it will be interesting, you will even get pleasure!

Methods and principles of finding partial derivatives of a function of three variables are actually very similar to partial derivative functions of two variables. The function of two variables, I remind you, has the form , where "x" and "y" are independent variables. Geometrically, a function of two variables is a certain surface in our three-dimensional space.

The function of three variables has the form , while the variables are called independentvariables or arguments, the variable is called dependent variable or function. For example: - a function of three variables

And now a little about science fiction films and aliens. You often hear about 4D, 5D, 10D, etc. spaces. Nonsense or not?
After all, the function of three variables implies the fact that all things take place in a four-dimensional space (indeed, there are four variables). The graph of a function of three variables is the so-called hypersurface. It is impossible to imagine it, since we live in a three-dimensional space (length/width/height). So that you are not bored with me, I offer a quiz. I will ask a few questions, and those who wish can try to answer them:

- Is there a fourth, fifth, etc. in the world? measurements in the sense of the philistine understanding of space (length/width/height)?

- Is it possible to build a four-dimensional, five-dimensional, etc. space in the broad sense of the word? That is, to give an example of such a space in our life.

Is it possible to travel to the past?

Is it possible to travel to the future?

- Do aliens exist?

For any question, you can choose one of four answers:
Yes / No (science forbids this) / Science does not forbid / Don't know

Whoever answers all the questions correctly, he most likely possesses some thing ;-)

I will gradually give out answers to questions during the lesson, do not skip the examples!

Actually, they flew. And now the good news: for a function of three variables, the rules of differentiation and the table of derivatives are valid. That is why you need to be good at managing the "ordinary" derivatives of functions one variable. There are very few differences!

Example 1

Solution: It is easy to guess that for a function of three variables there are three partial derivatives of the first order, which are denoted as follows:

Or - partial derivative of "x";
or - partial derivative with respect to "y";
or - partial derivative with respect to "z".

The notation with a stroke is more in use, but the compilers of collections, manuals in the conditions of tasks are very fond of using just cumbersome notations - so don't get lost! Perhaps not everyone knows how to read these "terrible fractions" aloud correctly. Example: should be read as follows: “de u po de x”.

Let's start with the x-derivative: . When we find the partial derivative with respect to , then the variables And are considered constants (constant numbers). And the derivative of any constant, oh, grace, is equal to zero:

Immediately pay attention to the subscript - no one forbids you to mark that they are constants. It’s even more convenient, I recommend that beginners use just such a record, there is less risk of confusion.

(1) We use the properties of the linearity of the derivative, in particular, we take out all the constants from the sign of the derivative. Please note that in the second term, the constant does not need to be taken out: since the “y” is a constant, then it is also a constant. In the term, the "usual" constant 8 and the constant "zet" are taken out of the sign of the derivative.

(2) We find the simplest derivatives, not forgetting that are constants. Next, comb the answer.

Partial derivative . When we find the partial derivative with respect to "y", then the variables And are considered constants:

(1) We use the properties of linearity. And again, note that the terms are constants, which means that nothing needs to be taken out for the sign of the derivative.

(2) We find derivatives, not forgetting that constants. Let's simplify the answer.

And finally, the partial derivative. When we find the partial derivative with respect to "z", then the variables And are considered constants:

General rule obvious and unpretentious: When we find the partial derivativefor any independent variable, thentwo others independent variables are considered constants.

When designing these tasks, you should be extremely careful, in particular, can't lose subscripts(which indicate on which variable differentiation is made). The loss of the index will be a GREAT FAULT. Hmmm…. it's funny if, after such an intimidation, I myself will miss them somewhere)

Example 2

Find partial derivatives of the first order of a function of three variables

This is a do-it-yourself example. Full solution and answer at the end of the lesson.

The two examples considered are quite simple and, having solved several similar problems, even a teapot will adapt to cracking down on them verbally.

To unload, let's return to the first question of the quiz: Is there a fourth, fifth, etc. in the world? measurements in the sense of the philistine understanding of space (length/width/height)?

Correct answer: Science doesn't forbid it.. All fundamental mathematical axiomatics, theorems, mathematical apparatus are beautiful and consistent work in space of any dimension. It is possible that somewhere in the Universe there are hypersurfaces that are not subject to our mind, for example, a four-dimensional hypersurface, which is given by a function of three variables. Or maybe there are hypersurfaces next to us or even we are right in them, just our vision, other sense organs, consciousness are capable of perceiving and comprehending only three dimensions.

Let's get back to the examples. Yes, if someone is heavily loaded with a quiz, it is better to read the answers to the following questions after you learn how to find the partial derivatives of a function of three variables, otherwise I will take out the whole brain for you in the course of the article =)

In addition to the simplest Examples 1,2, in practice there are tasks that can be called a small puzzle. Such examples, to my annoyance, fell out of sight when I created the lesson. Partial derivatives of functions of two variables. Making up for lost time:

Example 3

Solution: It seems to be “everything is simple”, but the first impression is deceptive. When finding partial derivatives, many will guess on the coffee grounds and make mistakes.

Let's analyze the example consistently, clearly and clearly.

Let's start with the partial derivative with respect to x. When we find the partial derivative with respect to "x", then the variables are considered constants. Therefore, the index of our function is also a constant. For dummies, I recommend the following solution: on the draft, change the constant to a specific positive integer, for example, to “five”. The result is a function of one variable:
or you can also write it like this:

This power function with complex base (sine). By :

Now remember that , thus:

On a clean copy, of course, the solution should be drawn up like this:

We find the partial derivative with respect to "y", they are considered constants. If "x" is a constant, then it is also a constant. On the draft, we do the same trick: we replace, for example, with 3, "Z" - we will replace it with the same "five". The result is again a function of one variable:

This demonstration function with a complex exponent. By the rule of differentiation of a complex function:

Now remember our replacement:

Thus:

On a clean copy, of course, the design should look nice:

And a mirror case with a partial derivative with respect to "z" (- constants):

With some experience, the analysis can be carried out mentally.

We carry out the second part of the task - we compose a differential of the first order. It is very simple, by analogy with a function of two variables, the first-order differential is written by the formula:

In this case:

And business then. I note that in practical problems, the full differential of the 1st order of a function of three variables is required to be compiled much less frequently than for a function of two variables.

A fun example for a do-it-yourself solution:

Example 4

Find first-order partial derivatives of a function of three variables and make a first-order total differential

Full solution and answer at the end of the lesson. If you have any difficulties, use the considered "chainikov's" algorithm, it is guaranteed to help. And another useful tip - do not hurry. Such examples are not quickly solved even by me.

We digress and analyze the second question: Is it possible to build a four-dimensional, five-dimensional, etc. space in the broad sense of the word? That is, to give an example of such a space in our life.

Correct answer: Yes. And, it's very easy. For example, we add a fourth dimension to the length/width/height - time. Popular four-dimensional space-time and the well-known theory of relativity carefully stolen by Einstein from Lobachevsky, Poincaré, Lorentz and Minkowski. Not everyone knows either. Why did Einstein get the Nobel Prize? There was a terrible scandal in the scientific world, and the Nobel Committee formulated the merit of the plagiarist as follows: "For the general contribution to the development of physics." So that's it. Einstein's C grade brand is pure promotion and PR.

It is easy to add a fifth dimension to the considered four-dimensional space, for example: atmospheric pressure. And so on, so on, so on, as many dimensions you set in your model - there will be so many. In the broad sense of the word, we live in a multidimensional space.

Let's look at a couple more typical tasks:

Example 5

Find first order partial derivatives at a point

Solution: A task in this formulation is often encountered in practice and involves the following two actions:
– you need to find partial derivatives of the first order;
– you need to calculate the values ​​of partial derivatives of the 1st order at the point .

We decide:

(1) We have a complex function, and the first step is to take the derivative of the arc tangent. In doing so, we, in fact, calmly use the tabular formula for the derivative of the arc tangent. By the rule of differentiation of a complex function the result must be multiplied by the derivative of the inner function (embedding): .

(2) We use the properties of linearity.

(3) And we take the remaining derivatives, not forgetting that they are constants.

According to the assignment condition, it is necessary to find the value of the found partial derivative at the point . Substitute the coordinates of the point in the found derivative:

The advantage of this task is the fact that other partial derivatives are found in a very similar way:

As you can see, the solution template is almost the same.

Let's calculate the value of the found partial derivative at the point :

And finally, the derivative with respect to "z":

Ready. The solution could also be formulated in another way: first, find all three partial derivatives, and then calculate their values ​​at the point . But, it seems to me, the above method is more convenient - they just found the partial derivative, and immediately, without leaving the cash register, calculated its value at a point.

It is interesting to note that, geometrically, a point is a very real point in our three-dimensional space. The values ​​of the function, derivatives are already the fourth dimension, and no one knows where it is geometrically located. As they say, no one crawled around the Universe with a tape measure, did not check.

As soon as the philosophical theme has gone again, let's consider the third question: Is it possible to travel into the past?

Correct answer: No. Traveling into the past contradicts the second law of thermodynamics about the irreversibility of physical processes (entropy). So please don't dive into a pool without water, the event can only be played back in the video =) Folk wisdom has come up with the opposite worldly law for a reason: "Measure seven times, cut once." Although, in fact, a sad thing, time is unidirectional and irreversible, none of us will look younger tomorrow. And various science fiction films like "Terminator" from a scientific point of view are complete nonsense. It is also absurd from the point of view of philosophy - when the Consequence, returning to the past, can destroy its own Cause. .

More interesting with the derivative with respect to "z", although, it's still almost the same:

(1) We take the constants out of the sign of the derivative.

(2) Here again the product of two functions, each of which depends from the "live" variable "z". In principle, you can use the formula for the derivative of a quotient, but it's easier to go the other way - to find the derivative of the product.

(3) A derivative is a tabular derivative. The second term contains the already familiar derivative of a complex function.

Example 9

Find partial derivatives of the first order of a function of three variables

This is a do-it-yourself example. Think about how it is more rational to find one or another partial derivative. Full solution and answer at the end of the lesson.

Before proceeding to the final examples of the lesson and consider second order partial derivatives functions of three variables, I will once again cheer everyone up with the fourth question:

Is it possible to travel to the future?

Correct answer: Science doesn't forbid it.. Paradoxically, there is no mathematical, physical, chemical or other natural science law that would prohibit travel to the future! Seems like nonsense? But almost everyone in life had a premonition (and not supported by any logical arguments) that this or that event would happen. And it happened! Where did the information come from? From the future? Thus, fantastic films about traveling into the future, and, by the way, the predictions of all kinds of fortune-tellers, psychics cannot be called such nonsense. At least, science has not refuted this. Everything is possible! So, when I was in school, CDs and flat-panel monitors from movies seemed like incredible fantasy to me.

The well-known comedy "Ivan Vasilyevich Changes His Profession" is half fiction (as a maximum). No scientific law forbade Ivan the Terrible to be in the future, but it is impossible for two peppers to be in the past and perform the duties of a king.