Examples of calculating higher order derivatives of explicit functions are considered. Useful formulas for calculating nth order derivatives are given.

Content

Determination of higher order derivatives

Here we consider the case where the variable y depends on the variable x explicitly:
.
Differentiating the function with respect to the variable x, we obtain the first-order derivative, or simply the derivative:
.
As a result, we obtain a new function, which is a derivative of the function. Differentiating this new function with respect to the variable x, we obtain the second-order derivative:
.
Differentiating the function, we obtain a third-order derivative:
.
And so on. Differentiating the original function n times, we obtain the nth order derivative or nth derivative:
.

Derivatives can be denoted strokes, Roman numerals, Arabic numerals in brackets or fractions from differentials. For example, derivatives of the third and fourth orders can be denoted as follows:
;
.

Below are formulas that may be useful in calculating higher order derivatives.

Useful formulas for nth order derivatives

Derivatives of some elementary functions :
;
;
;
;
.

Derivative of the sum of functions:
,
where are constants.

Leibniz formula derivative of the product of two functions:
,
Where
- binomial coefficients.

Example 1

Find the first and second order derivatives of the following function:
.

We find the first order derivative. We take the constant outside the derivative sign and apply the formula from the table of derivatives:
.
We apply the rule of differentiation of complex functions:
.
Here .
We apply the rule of differentiation of a complex function and use the found derivatives:
.
Here .

.
To find the second-order derivative, we need to find the derivative of the first-order derivative, that is, of the function:
.
To avoid confusion with the notation, let’s denote this function by the letter :
(A1.1) .
Then second order derivative from the original function is the derivative of the function:
.

Finding the derivative of the function. This is easier to do using the logarithmic derivative. Let's logarithmize (A1.1):
.
Now let's differentiate:
(A1.2) .
But it's constant. Its derivative is zero. We have already found the derivative of. We find the remaining derivatives using the rule of differentiation of a complex function.
;
;
.
We substitute in (A1.2):

.
From here
.

;
.

Example 2

Find the third order derivative:
.

Finding the first order derivative. To do this, we take the constant outside the sign of the derivative and use table of derivatives and apply rule for finding the derivative of a complex function .

.
Here .
So, we found the first order derivative:
.

Finding the second order derivative. To do this, we find the derivative of . We apply the derivative fraction formula.
.
Second order derivative:
.

Now we find what we are looking for third order derivative. To do this, we differentiate.
;
;

.

The third order derivative is equal to
.

Example 3

Find the sixth order derivative of the following function:
.

If you open the brackets, it will be clear that the original function is a polynomial of degree . Let's write it as a polynomial:
,
where are constant coefficients.

Next we apply nth formula derivative of a power function:
.
For the sixth order derivative (n = 6 ) we have:
.
From this it is clear that at . When we have:
.

We use the formula for the derivative of a sum of functions:

.
Thus, to find the sixth-order derivative of the original function, we only need to find the coefficient of the polynomial at the highest degree. We find it by multiplying the highest powers in the products of the sums of the original function:

.
From here. Then
.

Example 4

Find the nth derivative of a function
.

Solution > > >

Example 5

Find the nth derivative of the following function:
,
where and are constants.

In this example, it is convenient to perform calculations using complex numbers. Let us have some complex function
(A5.1) ,
where and are functions of the real variable x;
- imaginary unit, .
Differentiating (A.1) n times, we have:
(A5.2) .
Sometimes it's easier to find the nth derivative of a function. Then the nth derivatives of the functions are defined as the real and imaginary parts of the nth derivative:
;
.

Let's use this technique to solve our example. Consider the function
.
Here we have applied Euler's formula
,
and introduced the designation
.
Then the nth derivative of the original function is determined by the formula:
.

Let's find the nth derivative of the function
.
To do this we apply the formula:
.
In our case
.
Then
.

So, we found the nth derivative of the complex function:
,
Where .
Let's find the real part of the function.
To do this, let's imagine complex number in demonstrative form:
,
Where ;
; .
Then
;

.

Example solution
.

Let , .
Then ;
.
At ,
,
,
.
And we get the formula for the nth derivative of the cosine:
.

,
Where
; .

Consider a function of two variables:

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

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, you 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 other variables except the desired one, and then find the ordinary derivative with respect to this desired variable.

This leads to the main technique for calculating such derivatives: simply assume that all variables except this one are a constant, and then differentiate the function as you would differentiate an “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 derivative sign, and in the second case we completely zeroed out the first term. All this happens 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 "partial derivative"?

Today we will talk about functions of several variables and partial derivatives of them. First, what is a function of several variables? Until now, we are accustomed to consider a function as $y\left(x \right)$ or $t\left(x \right)$, or any variable and one single function of it. Now we will have one function, but several variables. As $y$ and $x$ change, the value of the function will change. For example, if $x$ doubles, the value of the function will change, and if $x$ changes, but $y$ 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 calculate the derivative of a function from any variable, we are required to indicate for which variable we are calculating the derivative - 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 for 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 to be a constant and treat it as such. In particular, when calculating the derivative of a product, we can take $y$ out of brackets (we have a constant), and when calculating the derivative of a sum, if somewhere we get a 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 a constant.

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

Problems with radicals and polynomials

Task No. 1

In order not to waste time, let's start from the very beginning with serious examples.

To begin with, let me remind you of this 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 another one of this expression with respect to the same variable. Let's first calculate 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 little:

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

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

The answer has been 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 it down 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 down:

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

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

Done.

Problem No. 2

This example is both simpler and more complex than the previous one. It’s more complicated because there are more actions, but it’s simpler 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 will not change. This remark will further simplify our calculation of the partial derivative, i.e. it is enough to count one of them, and in the second one simply 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 this notation, so let’s 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 calculate them, if all the rules are applied correctly, the answer will be the same.

Now let's look at one more partial derivative from our big formula:

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

Let's substitute the resulting expressions into our formula and get:

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

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

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

Based on $x$ counted. And to calculate $y$ from the same expression, let's not perform the same sequence of actions, but take advantage of 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 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 quotient. At the same time, however, specific features arise: if we consider the partial derivative of $x$, then when we obtain it from $x$, we consider it as a constant, and therefore its derivative will be equal to “zero”.

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

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

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

At the same time, 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 just polynomials and roots: there are also trigonometry, and logarithms, and the exponential function. Now let's do this.

Problems with trigonometric functions and logarithms

Task No. 1

Let us 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 out one variable separately:

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

Let's return to our design:

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

That's it, we found it 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, let's calculate 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.

Problem No. 2

Let's write down 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 for $x$. We count 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))\ ]

The problem is 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 of working with standard derivatives remain unchanged, namely, the derivative of a sum and a difference, a quotient and a complex function.

The last formula is most often found when solving problems with partial derivatives. We meet them almost everywhere. There has never been a single task where we didn’t come across it. But no matter what formula we use, we still have one more requirement added, namely, the peculiarity of working with partial derivatives. Once 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 $y$ is the variable, and $x$ remains constant everywhere. The same thing works the other way around. It can be taken out of the derivative sign, 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, let's look at 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 No. 1

To begin with, let's write 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 it 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 with 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, $x$ is calculated.

However, as I promised, now we will try to calculate this 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 received exactly the same answer, but the amount of calculations turned out to be smaller. To do this, it was enough to note that when performing the product, the indicators 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 solving 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, this same derivative could be calculated in the second way, and the answer would be the same.

Problem No. 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 calculate 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 solving the original construction: $$

This is the answer.

It remains to find by analogy using $y$:

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

As always, we calculate one expression separately:

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

We continue solving the basic design:

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

Nuances of the solution

Here shining example 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, i.e. differentiation, the answer may turn out 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 calculate two more examples.

Problems with trigonometric functions and functions with three variables

Task No. 1

Let's write down the following formulas:

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

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

Let's now solve our expression:

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

Let’s separately calculate 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 response of the private variable on $x$. 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:

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

Let's solve our construction to the end:

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

Problem No. 2

At first glance, this example may seem quite complicated because there are three variables. In fact, this is one of the most simple 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 all that remains is 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, which completes the solution to the second problem.

Nuances of the solution

As you can see, there is nothing complicated in these two examples. The only thing we are convinced of is that the derivative of a complex function is used often and depending on which partial derivative we calculate, 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 were convinced that they are all significantly different from each other.

Key points

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

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

Of course, watching this video lesson alone is not enough to fully understand this topic, so right now on my website there is a set of problems for this video specifically dedicated to today’s topic - go in, download, solve these problems and check the answer. And after that, no problems with partial derivatives either in exams or in independent work you won't have. Of course, this is not the last lesson in higher mathematics, so visit our website, add VKontakte, subscribe to YouTube, like and stay with us!

Let the function be given. Since x and y are independent variables, one of them can change while the other maintains its value. Let's give the independent variable x an increment while keeping the value of y unchanged. Then z will receive an increment, which is called the partial increment of z with respect to x and is denoted . So, .

Similarly, we obtain the partial increment of z over y: .

The total increment of the function z is determined by the equality .

If there is a limit, then it is called the partial derivative of the function at a point with respect to the variable x and is denoted by one of the symbols:

.

Partial derivatives with respect to x at a point are usually denoted by the symbols .

The partial derivative of with respect to the variable y is defined and denoted similarly:

Thus, the partial derivative of a function of several (two, three or more) variables is defined as the derivative of a function of one of these variables, provided that the values ​​of the remaining independent variables are constant. Therefore, partial derivatives of a function are found using the formulas and rules for calculating derivatives of a function of one variable (in this case, x or y are considered respectively constant value).

Partial derivatives are called first-order partial derivatives. They can be considered as functions of . These functions can have partial derivatives, which are called second-order partial derivatives. They are defined and labeled as follows:

; ;

; .

Differentials of 1st and 2nd order of a function of two variables.

Full differential functions (formula 2.5) are called first order differentials.

The formula for calculating the total differential is as follows:

(2.5) or , Where ,

partial differentials of a function.

Let the function have continuous partial derivatives of the second order. The second order differential is determined by the formula. Let's find it:

From here: . Symbolically it is written like this:

.

UNDETERMINED INTEGRAL.

Antiderivative of function, indefinite integral, properties.

The function F(x) is called antiderivative for a given function f(x), if F"(x)=f(x), or, what is the same, if dF(x)=f(x)dx.

Theorem. If a function f(x), defined in some interval (X) of finite or infinite length, has one antiderivative, F(x), then it also has infinitely many antiderivatives; all of them are contained in the expression F(x) + C, where C is an arbitrary constant.

The set of all antiderivatives for a given function f(x), defined in a certain interval or on a segment of finite or infinite length, is called indefinite integral from the function f(x) [or from the expression f(x)dx ] and is denoted by the symbol .

If F(x) is one of the antiderivatives for f(x), then according to the antiderivative theorem

, where C is an arbitrary constant.

By definition of an antiderivative, F"(x)=f(x) and, therefore, dF(x)=f(x) dx. In formula (7.1), f(x) is called an integrand function, and f(x) dx is called an integrand expression.

Partial derivatives of a function of two variables.
Concept and examples of solutions

In this lesson we will continue our acquaintance with the function of two variables and consider perhaps the most common thematic task - finding partial derivatives of the first and second order, as well as the total differential of the function. Part-time students, as a rule, encounter partial derivatives in the 1st year in the 2nd semester. Moreover, according to my observations, the task of finding partial derivatives almost always appears on the exam.

For effective learning the following material for you necessary be able to more or less confidently find “ordinary” derivatives of functions of one variable. You can learn how to handle derivatives correctly in lessons How to find the derivative? And Derivative of a complex function. We will also need a table of derivatives of elementary functions and differentiation rules; it is most convenient if it is at hand in printed form. Get it reference material possible on the page Mathematical formulas and tables.

Let's quickly repeat the concept of a function of two variables, I will try to limit myself to the bare minimum. A function of two variables is usually written as , with the variables being called independent variables or arguments.

Example: – function of two variables.

Sometimes the notation is used. There are also tasks where the letter is used instead of a letter.

WITH geometric point In terms of vision, a function of two variables most often represents a surface of three-dimensional space (plane, cylinder, sphere, paraboloid, hyperboloid, etc.). But, in fact, this is more analytic geometry, and on our agenda is mathematical analysis, which my university teacher never let me cheat on and is my strong point.

Let's move on to the question of finding partial derivatives of the first and second orders. I have some good news for those who have had a few cups of coffee and are tuning in to some incredibly difficult material: partial derivatives are almost the same as “ordinary” derivatives of a function of one variable.

For partial derivatives, all differentiation rules and the table of derivatives of elementary functions are valid. There are only a couple of small differences, which we will get to know right now:

...yes, by the way, for this topic I created small pdf book, which will allow you to “get your teeth into” in just a couple of hours. But by using the site, you will certainly get the same result - just maybe a little slower:

Example 1

Find the first and second order partial derivatives of the function

First, let's find the first-order partial derivatives. There are two of them.

Designations:
or – partial derivative with respect to “x”
or – partial derivative with respect to “y”

Let's start with . When we find the partial derivative with respect to “x”, the variable is considered a constant (constant number).

Comments on the actions performed:

(1) The first thing we do when finding the partial derivative is to conclude all function in brackets under the prime with subscript.

Attention, important! WE DO NOT LOSE subscripts during the solution process. In this case, if you draw a “stroke” somewhere without , then the teacher, at a minimum, can put it next to the assignment (immediately bite off part of the point for inattention).

(2) We use the rules of differentiation , . For simple example like this one, both rules can easily be applied in one step. Pay attention to the first term: since is considered a constant, and any constant can be taken out of the derivative sign, then we put it out of brackets. That is, in this situation it is no better than an ordinary number. Now let's look at the third term: here, on the contrary, there is nothing to take out. Since it is a constant, it is also a constant, and in this sense it is no better than the last term - “seven”.

(3) We use tabular derivatives and .

(4) Let’s simplify, or, as I like to say, “tweak” the answer.

Now . When we find the partial derivative with respect to “y”, then the variableconsidered a constant (constant number).

(1) We use the same differentiation rules , . In the first term we take the constant out of the sign of the derivative, in the second term we can’t take anything out since it is already a constant.

(2) We use the table of derivatives of elementary functions. Let’s mentally change all the “X’s” in the table to “I’s”. That is, this table is equally valid for (and indeed for almost any letter). In particular, the formulas we use look like this: and .

What is the meaning of partial derivatives?

In essence, 1st order partial derivatives resemble "ordinary" derivative:

- This functions, which characterize rate of change functions in the direction of the and axes, respectively. So, for example, the function characterizes the steepness of “rises” and “slopes” surfaces in the direction of the abscissa axis, and the function tells us about the “relief” of the same surface in the direction of the ordinate axis.

! Note : here we mean directions that parallel coordinate axes.

For the purpose of better understanding, let’s consider a specific point on the plane and calculate the value of the function (“height”) at it:
– and now imagine that you are here (ON THE surface).

Let's calculate the partial derivative with respect to "x" at a given point:

The negative sign of the “X” derivative tells us about decreasing functions at a point in the direction of the abscissa axis. In other words, if we make a small, small (infinitesimal) step towards the tip of the axis (parallel to this axis), then we will go down the slope of the surface.

Now we find out the nature of the “terrain” in the direction of the ordinate axis:

The derivative with respect to the “y” is positive, therefore, at a point in the direction of the axis the function increases. To put it simply, here we are waiting for an uphill climb.

In addition, the partial derivative at a point characterizes rate of change functions in the corresponding direction. The greater the resulting value modulo– the steeper the surface, and vice versa, the closer it is to zero, the flatter the surface. So, in our example, the “slope” in the direction of the abscissa axis is steeper than the “mountain” in the direction of the ordinate axis.

But those were two private paths. It is quite clear that from the point we are at, (and in general from any point on a given surface) we can move in some other direction. Thus, there is an interest in creating a general "navigation map" that would inform us about the "landscape" of the surface if possible at every point domain of definition of this function along all available paths. About this and others interesting things I’ll tell you in one of the next lessons, but for now let’s return to the technical side of the issue.

Let us systematize the elementary applied rules:

1) When we differentiate with respect to , the variable is considered a constant.

2) When differentiation is carried out according to, then is considered a constant.

3) The rules and table of derivatives of elementary functions are valid and applicable for any variable (or any other) by which differentiation is carried out.

Step two. We find second-order partial derivatives. There are four of them.

Designations:
or – second derivative with respect to “x”
or – second derivative with respect to “y”
or - mixed derivative of “x by igr”
or - mixed derivative of "Y"

There are no problems with the second derivative. Speaking in simple language, the second derivative is the derivative of the first derivative.

For convenience, I will rewrite the first-order partial derivatives already found:

First, let's find mixed derivatives:

As you can see, everything is simple: we take the partial derivative and differentiate it again, but in this case - this time according to the “Y”.

Likewise:

In practical examples, you can focus on the following equality:

Thus, through second-order mixed derivatives it is very convenient to check whether we have found the first-order partial derivatives correctly.

Find the second derivative with respect to “x”.
No inventions, let's take it and differentiate it by “x” again:

Likewise:

It should be noted that when finding, you need to show increased attention , since there are no miraculous equalities to verify them.

Second derivatives also find wide practical use, in particular, they are used in the task of finding extrema of a function of two variables. But everything has its time:

Example 2

Calculate the first order partial derivatives of the function at the point. Find second order derivatives.

This is an example for you to solve on your own (answers at the end of the lesson). If you have difficulty differentiating roots, return to the lesson How to find the derivative? In general, pretty soon you will learn to find such derivatives “on the fly.”

Let's get our hands on more complex examples:

Example 3

Check that . Write down the first order total differential.

Solution: Find the first order partial derivatives:

Pay attention to the subscript: , next to the “X” it is not forbidden to write in parentheses that it is a constant. This note can be very useful for beginners to make it easier to navigate the solution.

Further comments:

(1) We move all constants beyond the sign of the derivative. In this case, and , and, therefore, their product is considered a constant number.

(2) Don’t forget how to correctly differentiate roots.

(1) We take all constants out of the sign of the derivative; in this case, the constant is .

(2) Under the prime we have the product of two functions left, therefore, we need to use the rule for differentiating the product .

(3) Do not forget that this is a complex function (albeit the simplest of complex ones). We use the corresponding rule: .

Now we find mixed derivatives of the second order:

This means that all calculations were performed correctly.

Let's write down the total differential. In the context of the task under consideration, it makes no sense to tell what the total differential of a function of two variables is. It is important that this very differential very often needs to be written down in practical problems.

First order total differential function of two variables has the form:

In this case:

That is, you just need to stupidly substitute the already found first-order partial derivatives into the formula. In this and similar situations, it is best to write differential signs in numerators:

And according to repeated requests from readers, second order complete differential.

It looks like this:

Let's CAREFULLY find the “one-letter” derivatives of the 2nd order:

and write down the “monster”, carefully “attaching” the squares, the product and not forgetting to double the mixed derivative:

It's okay if something seems difficult; you can always come back to derivatives later, after you've mastered the differentiation technique:

Example 4

Find first order partial derivatives of a function . Check that . Write down the first order total differential.

Let's look at a series of examples with complex functions:

Example 5

Find the first order partial derivatives of the function.

Solution:

Example 6

Find first order partial derivatives of a function .
Write down the total differential.

This is an example for you to solve on your own (answer at the end of the lesson). I won't give you a complete solution because it's quite simple.

Quite often, all of the above rules are applied in combination.

Example 7

Find first order partial derivatives of a function .

(1) We use the rule for differentiating the sum

(2) The first term in this case is considered a constant, since there is nothing in the expression that depends on the “x” - only “y”. You know, it’s always nice when a fraction can be turned into zero). For the second term we apply the product differentiation rule. By the way, in this sense, nothing would have changed if a function had been given instead - the important thing is that here product of two functions, EACH of which depends on "X", and therefore, you need to use the product differentiation rule. For the third term, we apply the rule of differentiation of a complex function.

(1) The first term in both the numerator and denominator contains a “Y”, therefore, you need to use the rule for differentiating quotients: . The second term depends ONLY on “x”, which means it is considered a constant and turns to zero. For the third term we use the rule for differentiating a complex function.

For those readers who courageously made it almost to the end of the lesson, I’ll tell you an old Mekhmatov joke for relief:

One day, an evil derivative appeared in the space of functions and started to differentiate everyone. All functions are scattered in all directions, no one wants to transform! And only one function does not run away. The derivative approaches her and asks:

- Why don’t you run away from me?

- Ha. But I don’t care, because I am “e to the power of X”, and you won’t do anything to me!

To which the evil derivative with an insidious smile replies:

- This is where you are mistaken, I will differentiate you by “Y”, so you should be a zero.

Whoever understood the joke has mastered derivatives, at least to the “C” level).

Example 8

Find first order partial derivatives of a function .

This is an example for you to solve on your own. The complete solution and example of the problem are at the end of the lesson.

Well, that's almost all. Finally, I can’t help but please mathematics lovers with one more example. It's not even about amateurs, everyone has a different level of mathematical preparation - there are people (and not so rare) who like to compete with more difficult tasks. Although, the last example in this lesson is not so much complex as it is cumbersome from a computational point of view.

Let a function of two variables be given. Let's give the argument an increment and leave the argument unchanged. Then the function will receive an increment, which is called a partial increment by variable and is denoted:

Similarly, fixing the argument and giving an increment to the argument, we obtain a partial increment of the function by variable:

The quantity is called the total increment of the function at a point.

Definition 4. The partial derivative of a function of two variables with respect to one of these variables is the limit of the ratio of the corresponding partial increment of the function to the increment of a given variable when the latter tends to zero (if this limit exists). The partial derivative is denoted as follows: or, or.

Thus, by definition we have:

Partial derivatives of functions are calculated according to the same rules and formulas as a function of one variable, taking into account that when differentiating with respect to a variable, it is considered constant, and when differentiating with respect to a variable, it is considered constant.

Example 3. Find partial derivatives of functions:

Solution. a) To find, we consider it a constant value and differentiate it as a function of one variable:

Similarly, assuming a constant value, we find:

Definition 5. The total differential of a function is the sum of the products of the partial derivatives of this function by the increments of the corresponding independent variables, i.e.

Considering that the differentials of the independent variables coincide with their increments, i.e. , the formula for the total differential can be written as

Example 4. Find the complete differential of the function.

Solution. Since, using the total differential formula we find

Higher order partial derivatives

Partial derivatives are called first order partial derivatives or first partial derivatives.

Definition 6. The second-order partial derivatives of a function are the partial derivatives of the first-order partial derivatives.

There are four second order partial derivatives. They are designated as follows:

Partial derivatives of the 3rd, 4th and higher orders are defined similarly. For example, for a function we have:

Partial derivatives of the second or higher order, taken with respect to different variables, are called mixed partial derivatives. For a function, these are derivatives. Note that in the case when the mixed derivatives are continuous, then equality holds.

Example 5. Find second-order partial derivatives of a function

Solution. The first order partial derivatives for this function are found in Example 3:

Differentiating with respect to the variables x and y, we obtain