How to find the derivative of a number to a complex power. Derivatives of hyperbolic functions

With this video I begin a long series of lessons on derivatives. This lesson consists of several parts.

First of all, I will tell you what derivatives are and how to calculate them, but not in sophisticated academic language, but the way I understand it myself and how I explain it to my students. Secondly, we will consider the simplest rule for solving problems in which we will look for derivatives of sums, derivatives of differences and derivatives of a power function.

We will look at more complex combined examples, from which you will, in particular, learn that similar problems involving roots and even fractions can be solved using the formula for the derivative of a power function. In addition, of course, there will be many problems and examples of solutions of various levels of complexity.

In general, initially I was going to record a short 5-minute video, but you can see how it turned out. So enough of the lyrics - let's get down to business.

What is a derivative?

So, let's start from afar. Many years ago, when the trees were greener and life was more fun, mathematicians thought about this: consider a simple function defined by its graph, call it $y=f\left(x \right)$. Of course, the graph does not exist on its own, so you need to draw the $x$ axes as well as the $y$ axis. Now let's choose any point on this graph, absolutely any. Let's call the abscissa $((x)_(1))$, the ordinate, as you might guess, will be $f\left(((x)_(1)) \right)$.

Let's look at another point on the same graph. It doesn’t matter which one, the main thing is that it differs from the original one. It, again, has an abscissa, let's call it $((x)_(2))$, and also an ordinate - $f\left(((x)_(2)) \right)$.

So, we have two points: they have different abscissas and, therefore, different function values, although the latter is not necessary. But what is really important is that we know from the planimetry course: through two points you can draw a straight line and, moreover, only one. So let's carry it out.

Now let’s draw a straight line through the very first of them, parallel to the abscissa axis. We get a right triangle. Let's call it $ABC$, right angle $C$. This triangle has one very interesting property: the fact is that the angle $\alpha $ is actually equal to the angle at which the straight line $AB$ intersects with the continuation of the abscissa axis. Judge for yourself:

the straight line $AC$ is parallel to the $Ox$ axis by construction,
line $AB$ intersects $AC$ under $\alpha $,
hence $AB$ intersects $Ox$ under the same $\alpha $.

What can we say about $\text( )\!\!\alpha\!\!\text( )$? Nothing specific, except that in the triangle $ABC$ the ratio of leg $BC$ to leg $AC$ is equal to the tangent of this very angle. So let's write it down:

Of course, $AC$ in this case is easily calculated:

Likewise for $BC$:

In other words, we can write the following:

\[\operatorname(tg)\text( )\!\!\alpha\!\!\text( )=\frac(f\left(((x)_(2)) \right)-f\left( ((x)_(1)) \right))(((x)_(2))-((x)_(1)))\]

Now that we've got all that out of the way, let's go back to our chart and look at the new point $B$. Let's erase the old values and take $B$ somewhere closer to $((x)_(1))$. Let us again denote its abscissa by $((x)_(2))$, and its ordinate by $f\left(((x)_(2)) \right)$.

Let's look again at our little triangle $ABC$ and $\text( )\!\!\alpha\!\!\text( )$ inside it. It is quite obvious that this will be a completely different angle, the tangent will also be different because the lengths of the segments $AC$ and $BC$ have changed significantly, but the formula for the tangent of the angle has not changed at all - this is still the relationship between a change in the function and a change in the argument .

Finally, we continue to move $B$ closer to the original point $A$, as a result the triangle will become even smaller, and the straight line containing the segment $AB$ will look more and more like a tangent to the graph of the function.

As a result, if we continue to bring the points closer together, i.e., reduce the distance to zero, then the straight line $AB$ will indeed turn into a tangent to the graph at a given point, and $\text( )\!\!\alpha\!\ !\text( )$ will transform from a regular triangle element to the angle between the tangent to the graph and the positive direction of the $Ox$ axis.

And here we smoothly move on to the definition of $f$, namely, the derivative of a function at the point $((x)_(1))$ is the tangent of the angle $\alpha $ between the tangent to the graph at the point $((x)_( 1))$ and the positive direction of the $Ox$ axis:

\[(f)"\left(((x)_(1)) \right)=\operatorname(tg)\text( )\!\!\alpha\!\!\text( )\]

Returning to our graph, it should be noted that any point on the graph can be chosen as $((x)_(1))$. For example, with the same success we could remove the stroke at the point shown in the figure.

Let's call the angle between the tangent and the positive direction of the axis $\beta$. Accordingly, $f$ in $((x)_(2))$ will be equal to the tangent of this angle $\beta $.

\[(f)"\left(((x)_(2)) \right)=tg\text( )\!\!\beta\!\!\text( )\]

Each point on the graph will have its own tangent, and, therefore, its own function value. In each of these cases, in addition to the point at which we are looking for the derivative of a difference or sum, or the derivative of a power function, it is necessary to take another point located at some distance from it, and then direct this point to the original one and, of course, find out how in the process Such movement will change the tangent of the angle of inclination.

Derivative of a power function

Unfortunately, such a definition does not suit us at all. All these formulas, pictures, angles do not give us the slightest idea of how to calculate the real derivative in real problems. Therefore, let's digress a little from the formal definition and consider more effective formulas and techniques with which you can already solve real problems.

Let's start with the simplest constructions, namely, functions of the form $y=((x)^(n))$, i.e. power functions. In this case, we can write the following: $(y)"=n\cdot ((x)^(n-1))$. In other words, the degree that was in the exponent is shown in the front multiplier, and the exponent itself is reduced by unit. For example:

\[\begin(align)& y=((x)^(2)) \\& (y)"=2\cdot ((x)^(2-1))=2x \\\end(align) \]

Here's another option:

\[\begin(align)& y=((x)^(1)) \\& (y)"=((\left(x \right))^(\prime ))=1\cdot ((x )^(0))=1\cdot 1=1 \\& ((\left(x \right))^(\prime ))=1 \\\end(align)\]

Using these simple rules, let's try to remove the touch of the following examples:

So we get:

\[((\left(((x)^(6)) \right))^(\prime ))=6\cdot ((x)^(5))=6((x)^(5)) \]

Now let's solve the second expression:

\[\begin(align)& f\left(x \right)=((x)^(100)) \\& ((\left(((x)^(100)) \right))^(\ prime ))=100\cdot ((x)^(99))=100((x)^(99)) \\\end(align)\]

Of course, these were very simple tasks. However, real problems are more complex and they are not limited to just degrees of function.

So, rule No. 1 - if a function is presented in the form of the other two, then the derivative of this sum is equal to the sum of the derivatives:

\[((\left(f+g \right))^(\prime ))=(f)"+(g)"\]

Similarly, the derivative of the difference of two functions is equal to the difference of the derivatives:

\[((\left(f-g \right))^(\prime ))=(f)"-(g)"\]

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

In addition, there is another important rule: if some $f$ is preceded by a constant $c$, by which this function is multiplied, then the $f$ of this entire construction is calculated as follows:

\[((\left(c\cdot f \right))^(\prime ))=c\cdot (f)"\]

\[((\left(3((x)^(3)) \right))^(\prime ))=3((\left(((x)^(3)) \right))^(\ prime ))=3\cdot 3((x)^(2))=9((x)^(2))\]

Finally, one more very important rule: in problems there is often a separate term that does not contain $x$ at all. For example, we can observe this in our expressions today. The derivative of a constant, i.e., a number that does not depend in any way on $x$, is always equal to zero, and it does not matter at all what the constant $c$ is equal to:

\[((\left(c \right))^(\prime ))=0\]

Example solution:

\[((\left(1001 \right))^(\prime ))=((\left(\frac(1)(1000) \right))^(\prime ))=0\]

Key points again:

The derivative of the sum of two functions is always equal to the sum of the derivatives: $((\left(f+g \right))^(\prime ))=(f)"+(g)"$;
For similar reasons, the derivative of the difference of two functions is equal to the difference of two derivatives: $((\left(f-g \right))^(\prime ))=(f)"-(g)"$;
If a function has a constant factor, then this constant can be taken out as a derivative sign: $((\left(c\cdot f \right))^(\prime ))=c\cdot (f)"$;
If the entire function is a constant, then its derivative is always zero: $((\left(c \right))^(\prime ))=0$.

Let's see how it all works with real examples. So:

We write down:

\[\begin(align)& ((\left(((x)^(5))-3((x)^(2))+7 \right))^(\prime ))=((\left (((x)^(5)) \right))^(\prime ))-((\left(3((x)^(2)) \right))^(\prime ))+(7) "= \\& =5((x)^(4))-3((\left(((x)^(2)) \right))^(\prime ))+0=5((x) ^(4))-6x \\\end(align)\]

In this example we see both the derivative of the sum and the derivative of the difference. In total, the derivative is equal to $5((x)^(4))-6x$.

Let's move on to the second function:

Let's write down the solution:

\[\begin(align)& ((\left(3((x)^(2))-2x+2 \right))^(\prime ))=((\left(3((x)^( 2)) \right))^(\prime ))-((\left(2x \right))^(\prime ))+(2)"= \\& =3((\left(((x) ^(2)) \right))^(\prime ))-2(x)"+0=3\cdot 2x-2\cdot 1=6x-2 \\\end(align)\]

Here we have found the answer.

Let's move on to the third function - it is more serious:

\[\begin(align)& ((\left(2((x)^(3))-3((x)^(2))+\frac(1)(2)x-5 \right)) ^(\prime ))=((\left(2((x)^(3)) \right))^(\prime ))-((\left(3((x)^(2)) \right ))^(\prime ))+((\left(\frac(1)(2)x \right))^(\prime ))-(5)"= \\& =2((\left(( (x)^(3)) \right))^(\prime ))-3((\left(((x)^(2)) \right))^(\prime ))+\frac(1) (2)\cdot (x)"=2\cdot 3((x)^(2))-3\cdot 2x+\frac(1)(2)\cdot 1=6((x)^(2)) -6x+\frac(1)(2) \\\end(align)\]

We have found the answer.

Let's move on to the last expression - the most complex and longest:

So, we consider:

\[\begin(align)& ((\left(6((x)^(7))-14((x)^(3))+4x+5 \right))^(\prime ))=( (\left(6((x)^(7)) \right))^(\prime ))-((\left(14((x)^(3)) \right))^(\prime )) +((\left(4x \right))^(\prime ))+(5)"= \\& =6\cdot 7\cdot ((x)^(6))-14\cdot 3((x )^(2))+4\cdot 1+0=42((x)^(6))-42((x)^(2))+4 \\\end(align)\]

But the solution does not end there, because we are asked not just to remove a stroke, but to calculate its value at a specific point, so we substitute −1 instead of $x$ into the expression:

\[(y)"\left(-1 \right)=42\cdot 1-42\cdot 1+4=4\]

Let's go further and move on to even more complex and interesting examples. The fact is that the formula for solving the power derivative $((\left(((x)^(n)) \right))^(\prime ))=n\cdot ((x)^(n-1))$ has an even wider scope than is usually believed. With its help, you can solve examples with fractions, roots, etc. This is what we will do now.

To begin with, let’s once again write down the formula that will help us find the derivative of a power function:

And now attention: so far we have considered only natural numbers as $n$, but nothing prevents us from considering fractions and even negative numbers. For example, we can write the following:

\[\begin(align)& \sqrt(x)=((x)^(\frac(1)(2))) \\& ((\left(\sqrt(x) \right))^(\ prime ))=((\left(((x)^(\frac(1)(2))) \right))^(\prime ))=\frac(1)(2)\cdot ((x) ^(-\frac(1)(2)))=\frac(1)(2)\cdot \frac(1)(\sqrt(x))=\frac(1)(2\sqrt(x)) \\\end(align)\]

Nothing complicated, so let's see how this formula will help us when solving more complex problems. So, an example:

Let's write down the solution:

\[\begin(align)& \left(\sqrt(x)+\sqrt(x)+\sqrt(x) \right)=((\left(\sqrt(x) \right))^(\prime ))+((\left(\sqrt(x) \right))^(\prime ))+((\left(\sqrt(x) \right))^(\prime )) \\& ((\ left(\sqrt(x) \right))^(\prime ))=\frac(1)(2\sqrt(x)) \\& ((\left(\sqrt(x) \right))^( \prime ))=((\left(((x)^(\frac(1)(3))) \right))^(\prime ))=\frac(1)(3)\cdot ((x )^(-\frac(2)(3)))=\frac(1)(3)\cdot \frac(1)(\sqrt(((x)^(2)))) \\& (( \left(\sqrt(x) \right))^(\prime ))=((\left(((x)^(\frac(1)(4))) \right))^(\prime )) =\frac(1)(4)((x)^(-\frac(3)(4)))=\frac(1)(4)\cdot \frac(1)(\sqrt(((x) ^(3)))) \\\end(align)\]

Let's go back to our example and write:

\[(y)"=\frac(1)(2\sqrt(x))+\frac(1)(3\sqrt(((x)^(2))))+\frac(1)(4 \sqrt(((x)^(3))))\]

This is such a difficult decision.

Let's move on to the second example - there are only two terms, but each of them contains both a classical degree and roots.

Now we will learn how to find the derivative of a power function, which, in addition, contains the root:

\[\begin(align)& ((\left(((x)^(3))\sqrt(((x)^(2)))+((x)^(7))\sqrt(x) \right))^(\prime ))=((\left(((x)^(3))\cdot \sqrt(((x)^(2))) \right))^(\prime )) =((\left(((x)^(3))\cdot ((x)^(\frac(2)(3))) \right))^(\prime ))= \\& =(( \left(((x)^(3+\frac(2)(3))) \right))^(\prime ))=((\left(((x)^(\frac(11)(3 ))) \right))^(\prime ))=\frac(11)(3)\cdot ((x)^(\frac(8)(3)))=\frac(11)(3)\ cdot ((x)^(2\frac(2)(3)))=\frac(11)(3)\cdot ((x)^(2))\cdot \sqrt(((x)^(2 ))) \\& ((\left(((x)^(7))\cdot \sqrt(x) \right))^(\prime ))=((\left(((x)^(7 ))\cdot ((x)^(\frac(1)(3))) \right))^(\prime ))=((\left(((x)^(7\frac(1)(3 ))) \right))^(\prime ))=7\frac(1)(3)\cdot ((x)^(6\frac(1)(3)))=\frac(22)(3 )\cdot ((x)^(6))\cdot \sqrt(x) \\\end(align)\]

Both terms have been calculated, all that remains is to write down the final answer:

\[(y)"=\frac(11)(3)\cdot ((x)^(2))\cdot \sqrt(((x)^(2)))+\frac(22)(3) \cdot ((x)^(6))\cdot \sqrt(x)\]

We have found the answer.

Derivative of a fraction through a power function

But the possibilities of the formula for solving the derivative of a power function do not end there. The fact is that with its help you can calculate not only examples with roots, but also with fractions. This is precisely the rare opportunity that greatly simplifies the solution of such examples, but is often ignored not only by students, but also by teachers.

So, now we will try to combine two formulas at once. On the one hand, the classical derivative of a power function

\[((\left(((x)^(n)) \right))^(\prime ))=n\cdot ((x)^(n-1))\]

On the other hand, we know that an expression of the form $\frac(1)(((x)^(n)))$ can be represented as $((x)^(-n))$. Hence,

\[\left(\frac(1)(((x)^(n))) \right)"=((\left(((x)^(-n)) \right))^(\prime ) )=-n\cdot ((x)^(-n-1))=-\frac(n)(((x)^(n+1)))\]

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

Thus, the derivatives of simple fractions, where the numerator is a constant and the denominator is a degree, are also calculated using the classical formula. Let's see how this works in practice.

So, the first function:

\[((\left(\frac(1)(((x)^(2))) \right))^(\prime ))=((\left(((x)^(-2)) \ right))^(\prime ))=-2\cdot ((x)^(-3))=-\frac(2)(((x)^(3)))\]

The first example is solved, let's move on to the second:

\[\begin(align)& ((\left(\frac(7)(4((x)^(4)))-\frac(2)(3((x)^(3)))+\ frac(5)(2)((x)^(2))+2((x)^(3))-3((x)^(4)) \right))^(\prime ))= \ \& =((\left(\frac(7)(4((x)^(4))) \right))^(\prime ))-((\left(\frac(2)(3(( x)^(3))) \right))^(\prime ))+((\left(2((x)^(3)) \right))^(\prime ))-((\left( 3((x)^(4)) \right))^(\prime )) \\& ((\left(\frac(7)(4((x)^(4))) \right))^ (\prime ))=\frac(7)(4)((\left(\frac(1)(((x)^(4))) \right))^(\prime ))=\frac(7 )(4)\cdot ((\left(((x)^(-4)) \right))^(\prime ))=\frac(7)(4)\cdot \left(-4 \right) \cdot ((x)^(-5))=\frac(-7)(((x)^(5))) \\& ((\left(\frac(2)(3((x)^ (3))) \right))^(\prime ))=\frac(2)(3)\cdot ((\left(\frac(1)(((x)^(3))) \right) )^(\prime ))=\frac(2)(3)\cdot ((\left(((x)^(-3)) \right))^(\prime ))=\frac(2)( 3)\cdot \left(-3 \right)\cdot ((x)^(-4))=\frac(-2)(((x)^(4))) \\& ((\left( \frac(5)(2)((x)^(2)) \right))^(\prime ))=\frac(5)(2)\cdot 2x=5x \\& ((\left(2 ((x)^(3)) \right))^(\prime ))=2\cdot 3((x)^(2))=6((x)^(2)) \\& ((\ left(3((x)^(4)) \right))^(\prime ))=3\cdot 4((x)^(3))=12((x)^(3)) \\\ end(align)\]...

Now we collect all these terms into a single formula:

\[(y)"=-\frac(7)(((x)^(5)))+\frac(2)(((x)^(4)))+5x+6((x)^ (2))-12((x)^(3))\]

We have received an answer.

However, before moving on, I would like to draw your attention to the form of writing the original expressions themselves: in the first expression we wrote $f\left(x \right)=...$, in the second: $y=...$ Many students get lost when they see different forms of recording. What is the difference between $f\left(x \right)$ and $y$? Nothing really. They are just different entries with the same meaning. It’s just that when we say $f\left(x \right)$, we are talking, first of all, about a function, and when we talk about $y$, we most often mean the graph of a function. Otherwise, this is the same thing, i.e., the derivative in both cases is considered the same.

Complex problems with derivatives

In conclusion, I would like to consider a couple of complex combined problems that use everything we have considered today. They contain roots, fractions, and sums. However, these examples will only be complex in today’s video tutorial, because truly complex derivative functions will be waiting for you ahead.

So, the final part of today's video lesson, consisting of two combined tasks. Let's start with the first of them:

\[\begin(align)& ((\left(((x)^(3))-\frac(1)(((x)^(3)))+\sqrt(x) \right))^ (\prime ))=((\left(((x)^(3)) \right))^(\prime ))-((\left(\frac(1)(((x)^(3) )) \right))^(\prime ))+\left(\sqrt(x) \right) \\& ((\left(((x)^(3)) \right))^(\prime ) )=3((x)^(2)) \\& ((\left(\frac(1)(((x)^(3))) \right))^(\prime ))=((\ left(((x)^(-3)) \right))^(\prime ))=-3\cdot ((x)^(-4))=-\frac(3)(((x)^ (4))) \\& ((\left(\sqrt(x) \right))^(\prime ))=((\left(((x)^(\frac(1)(3))) \right))^(\prime ))=\frac(1)(3)\cdot \frac(1)(((x)^(\frac(2)(3))))=\frac(1) (3\sqrt(((x)^(2)))) \\\end(align)\]

The derivative of the function is equal to:

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

The first example is solved. Let's consider the second problem:

In the second example we proceed similarly:

\[((\left(-\frac(2)(((x)^(4)))+\sqrt(x)+\frac(4)(x\sqrt(((x)^(3)) )) \right))^(\prime ))=((\left(-\frac(2)(((x)^(4))) \right))^(\prime ))+((\left (\sqrt(x) \right))^(\prime ))+((\left(\frac(4)(x\cdot \sqrt(((x)^(3)))) \right))^ (\prime ))\]

Let's calculate each term separately:

\[\begin(align)& ((\left(-\frac(2)(((x)^(4))) \right))^(\prime ))=-2\cdot ((\left( ((x)^(-4)) \right))^(\prime ))=-2\cdot \left(-4 \right)\cdot ((x)^(-5))=\frac(8 )(((x)^(5))) \\& ((\left(\sqrt(x) \right))^(\prime ))=((\left(((x)^(\frac( 1)(4))) \right))^(\prime ))=\frac(1)(4)\cdot ((x)^(-\frac(3)(4)))=\frac(1 )(4\cdot ((x)^(\frac(3)(4))))=\frac(1)(4\sqrt(((x)^(3)))) \\& ((\ left(\frac(4)(x\cdot \sqrt(((x)^(3)))) \right))^(\prime ))=((\left(\frac(4)(x\cdot ((x)^(\frac(3)(4)))) \right))^(\prime ))=((\left(\frac(4)(((x)^(1\frac(3 )(4)))) \right))^(\prime ))=4\cdot ((\left(((x)^(-1\frac(3)(4))) \right))^( \prime ))= \\& =4\cdot \left(-1\frac(3)(4) \right)\cdot ((x)^(-2\frac(3)(4)))=4 \cdot \left(-\frac(7)(4) \right)\cdot \frac(1)(((x)^(2\frac(3)(4))))=\frac(-7) (((x)^(2))\cdot ((x)^(\frac(3)(4))))=-\frac(7)(((x)^(2))\cdot \sqrt (((x)^(3)))) \\\end(align)\]

All terms have been calculated. Now we return to the original formula and add all three terms together. We get that the final answer will be like this:

\[(y)"=\frac(8)(((x)^(5)))+\frac(1)(4\sqrt(((x)^(3))))-\frac(7 )(((x)^(2))\cdot \sqrt(((x)^(3))))\]

And that is all. This was our first lesson. In the following lessons we will look at more complex constructions, and also find out why derivatives are needed in the first place.

Proof and derivation of the formulas for the derivative of the exponential (e to the x power) and the exponential function (a to the x power). Examples of calculating derivatives of e^2x, e^3x and e^nx. Formulas for derivatives of higher orders.

The derivative of an exponent is equal to the exponent itself (the derivative of e to the x power is equal to e to the x power):
(1) (e x )′ = e x.

The derivative of an exponential function with a base a is equal to the function itself multiplied by the natural logarithm of a:
(2) .

Derivation of the formula for the derivative of the exponential, e to the x power

An exponential is an exponential function whose base is equal to the number e, which is the following limit:
.
Here it can be either a natural number or a real number. Next, we derive formula (1) for the derivative of the exponential.

Derivation of the exponential derivative formula

Consider the exponential, e to the x power:
y = e x .
This function is defined for everyone. Let's find its derivative with respect to the variable x. By definition, the derivative is the following limit:
(3) .

Let's transform this expression to reduce it to known mathematical properties and rules. To do this we need the following facts:
A) Exponent property:
(4) ;
B) Property of logarithm:
(5) ;
IN) Continuity of the logarithm and the property of limits for a continuous function:
(6) .
Here is a function that has a limit and this limit is positive.
G) The meaning of the second remarkable limit:
(7) .

Let's apply these facts to our limit (3). We use property (4):
;
.

Let's make a substitution. Then ; .
Due to the continuity of the exponential,
.
Therefore, when , . As a result we get:
.

Let's make a substitution. Then . At , . And we have:
.

Let's apply the logarithm property (5):
. Then
.

Let us apply property (6). Since there is a positive limit and the logarithm is continuous, then:
.
Here we also used the second remarkable limit (7). Then
.

Thus, we obtained formula (1) for the derivative of the exponential.

Derivation of the formula for the derivative of an exponential function

Now we derive formula (2) for the derivative of the exponential function with a base of degree a. We believe that and . Then the exponential function
(8)
Defined for everyone.

Let's transform formula (8). For this we will use properties of the exponential function and logarithm.
;
.
So, we transformed formula (8) to the following form:
.

Higher order derivatives of e to the x power

Now let's find derivatives of higher orders. Let's look at the exponent first:
(14) .
(1) .

We see that the derivative of function (14) is equal to function (14) itself. Differentiating (1), we obtain derivatives of the second and third order:
;
.

This shows that the nth order derivative is also equal to the original function:
.

Higher order derivatives of the exponential function

Now consider an exponential function with a base of degree a:
.
We found its first-order derivative:
(15) .

Differentiating (15), we obtain derivatives of the second and third order:
;
.

We see that each differentiation leads to the multiplication of the original function by . Therefore, the nth order derivative has the following form:
.

Definition of power-exponential function. Deriving a formula for calculating its derivative. Examples of calculating derivatives of power-exponential functions are analyzed in detail.

Power-exponential function is a function that has the form of a power function
y = u v ,
in which the base u and the exponent v are some functions of the variable x:
u = u (x); v = v (x).
This function is also called exponential or .

Note that the power-exponential function can be represented in exponential form:
.
Therefore it is also called complex exponential function.

Calculation using logarithmic derivative

Let's find the derivative of the power-exponential function
(2) ,
where and are functions of the variable.
To do this, we logarithm equation (2), using the property of the logarithm:
.
Differentiate with respect to the variable x:
(3) .
We apply rules for differentiating complex functions and works:
;
.

We substitute in (3):
.
From here
.

So, we found the derivative of the power-exponential function:
(1) .
If the exponent is constant, then . Then the derivative is equal to the derivative of a complex power function:
.
If the base of the degree is constant, then . Then the derivative is equal to the derivative of a complex exponential function:
.
When and are functions of x, then the derivative of the power-exponential function is equal to the sum of the derivatives of the complex power and exponential functions.

Calculation of the derivative by reduction to a complex exponential function

Now let's find the derivative of the power-exponential function
(2) ,
presenting it as a complex exponential function:
(4) .

Let's differentiate the product:
.
We apply the rule for finding the derivative of a complex function:

.
And we again got formula (1).

Example 1

Find the derivative of the following function:
.

Solution

We calculate using the logarithmic derivative. Let's logarithm the original function:
(A1.1) .

From the table of derivatives we find:
;
.
Using the product derivative formula, we have:
.
We differentiate (A1.1):
.
Because the
,
That
.

Answer

Example 2

Find the derivative of the function
.

Solution

Let's logarithm the original function:
(A2.1) .

Derivative calculation- one of the most important operations in differential calculus. Below is a table for finding derivatives of simple functions. For more complex differentiation rules, see other lessons:

Table of derivatives of exponential and logarithmic functions

Use the given formulas as reference values. They will help in solving differential equations and problems. In the picture, in the table of derivatives of simple functions, there is a “cheat sheet” of the main cases of finding a derivative in a form that is understandable for use, next to it are explanations for each case.

Derivatives of simple functions

1. The derivative of a number is zero
с´ = 0
Example:
5´ = 0

Explanation:
The derivative shows the rate at which the value of a function changes when its argument changes. Since the number does not change in any way under any conditions, the rate of its change is always zero.

2. Derivative of a variable equal to one
x´ = 1

Explanation:
With each increment of the argument (x) by one, the value of the function (the result of the calculation) increases by the same amount. Thus, the rate of change in the value of the function y = x is exactly equal to the rate of change in the value of the argument.

3. The derivative of a variable and a factor is equal to this factor
сx´ = с
Example:
(3x)´ = 3
(2x)´ = 2
Explanation:
In this case, every time the function argument changes ( X) its value (y) increases in With once. Thus, the rate of change of the function value in relation to the rate of change of the argument is exactly equal to the value With.

Whence it follows that
(cx + b)" = c
that is, the differential of the linear function y=kx+b is equal to the slope of the line (k).

4. Modulo derivative of a variable equal to the quotient of this variable to its modulus
|x|"= x / |x| provided that x ≠ 0
Explanation:
Since the derivative of a variable (see formula 2) is equal to one, the derivative of the module differs only in that the value of the rate of change of the function changes to the opposite when crossing the point of origin (try drawing a graph of the function y = |x| and see for yourself. This is exactly what value and returns the expression x / |x|. When x< 0 оно равно (-1), а когда x >0 - one. That is, for negative values of the variable x, with each increase in the argument, the value of the function decreases by exactly the same value, and for positive values, on the contrary, it increases, but by exactly the same value.

5. Derivative of a variable to a power equal to the product of a number of this power and a variable to the power reduced by one
(x c)"= cx c-1, provided that x c and cx c-1 are defined and c ≠ 0
Example:
(x 2)" = 2x
(x 3)" = 3x 2
To remember the formula:
Move the degree of the variable down as a factor, and then reduce the degree itself by one. For example, for x 2 - the two was ahead of the x, and then the reduced power (2-1 = 1) simply gave us 2x. The same thing happened for x 3 - we “move down” the triple, reduce it by one and instead of a cube we have a square, that is, 3x 2. A little "unscientific" but very easy to remember.

6.Derivative of a fraction 1/x
(1/x)" = - 1 / x 2
Example:
Since a fraction can be represented as raising to a negative power
(1/x)" = (x -1)", then you can apply the formula from rule 5 of the table of derivatives
(x -1)" = -1x -2 = - 1 / x 2

7. Derivative of a fraction with a variable of arbitrary degree in the denominator
(1 / x c)" = - c / x c+1
Example:
(1 / x 2)" = - 2 / x 3

8. Derivative of the root(derivative of variable under square root)
(√x)" = 1 / (2√x) or 1/2 x -1/2
Example:
(√x)" = (x 1/2)" means you can apply the formula from rule 5
(x 1/2)" = 1/2 x -1/2 = 1 / (2√x)

9. Derivative of a variable under the root of an arbitrary degree
(n √x)" = 1 / (n n √x n-1)

Derivation of the formula for the derivative of a power function (x to the power of a). Derivatives from roots of x are considered. Formula for the derivative of a higher order power function. Examples of calculating derivatives.

The derivative of x to the power of a is equal to a times x to the power of a minus one:
(1) .

The derivative of the nth root of x to the mth power is:
(2) .

Derivation of the formula for the derivative of a power function

Case x > 0

Consider a power function of the variable x with exponent a:
(3) .
Here a is an arbitrary real number. Let's first consider the case.

To find the derivative of function (3), we use the properties of a power function and transform it to the following form:
.

Now we find the derivative using:
;
.
Here .

Formula (1) has been proven.

Derivation of the formula for the derivative of a root of degree n of x to the degree of m

Now consider a function that is the root of the following form:
(4) .

To find the derivative, we transform the root to a power function:
.
Comparing with formula (3) we see that
.
Then
.

Using formula (1) we find the derivative:
(1) ;
;
(2) .

In practice, there is no need to memorize formula (2). It is much more convenient to first transform the roots to power functions, and then find their derivatives using formula (1) (see examples at the end of the page).

Case x = 0

If , then the power function is defined for the value of the variable x = 0 . Let's find the derivative of function (3) at x = 0 . To do this, we use the definition of a derivative:
.

Let's substitute x = 0 :
.
In this case, by derivative we mean the right-hand limit for which .

So we found:
.
From this it is clear that for , .
At , .
At , .
This result is also obtained from formula (1):
(1) .
Therefore, formula (1) is also valid for x = 0 .

Case x< 0

Consider function (3) again:
(3) .
For certain values of the constant a, it is also defined for negative values of the variable x. Namely, let a be a rational number. Then it can be represented as an irreducible fraction:
,
where m and n are integers that do not have a common divisor.

If n is odd, then the power function is also defined for negative values of the variable x. For example, when n = 3 and m = 1 we have the cube root of x:
.
It is also defined for negative values of the variable x.

Let us find the derivative of the power function (3) for and for rational values of the constant a for which it is defined. To do this, let's represent x in the following form:
.
Then ,
.
We find the derivative by placing the constant outside the sign of the derivative and applying the rule for differentiating a complex function:

.
Here . But
.
Since then
.
Then
.
That is, formula (1) is also valid for:
(1) .

Higher order derivatives

Now let's find higher order derivatives of the power function
(3) .
We have already found the first order derivative:
.

Taking the constant a outside the sign of the derivative, we find the second-order derivative:
.
Similarly, we find derivatives of the third and fourth orders:
;

.

From this it is clear that derivative of arbitrary nth order has the following form:
.

notice, that if a is a natural number, then the nth derivative is constant:
.
Then all subsequent derivatives are equal to zero:
,
at .

Examples of calculating derivatives

Example

Find the derivative of the function:
.

Solution

Let's convert roots to powers:
;
.
Then the original function takes the form:
.

Finding derivatives of powers:
;
.
The derivative of the constant is zero:
.