Proof and derivation of the formulas for the derivative of the exponential (e to the power of x) and 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.

Content

See also: Exponential function - properties, formulas, graph
Exponent, e to the x power - properties, formulas, graph

Basic formulas

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) .

An exponential is an exponential function whose power 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.
(3) .

Let's find its derivative with respect to the variable x.
By definition, the derivative is the following limit: Let's transform this expression to reduce it to known mathematical properties and rules. To do this we need the following facts:
(4) ;
A) Exponent property:
(5) ;
B) Property of logarithm:
(6) .
IN)
Continuity of the logarithm and the property of limits for a continuous function: Here is a function that has a limit and this limit is positive.
(7) .

G)
;
.

The meaning of the second remarkable limit:
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):

.
(8)
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.

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:

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 simple function

, given by its graph, let's 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 got two points: they have different abscissas and, therefore, functions, although the latter is optional. 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 right triangle. Let's call it $ABC$, right angle $C$. This triangle has one very interesting property: the fact is that the angle $\alpha $, in fact, equal to angle, under 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 $((x)_(1))$ can be any point on the graph. 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 one more 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 on 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 doesn’t 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 tasks. 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 shapes records. 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:

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

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 count 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 cover more complex designs, and also find out why derivatives are needed at all.

Very easy to remember.

Well, let’s not go far, let’s immediately consider the inverse function. Which function is the inverse of the exponential function? Logarithm:

In our case, the base is the number:

Such a logarithm (that is, a logarithm with a base) is called “natural”, and we use a special notation for it: we write instead.

What is it equal to? Of course, .

Derivative of natural logarithm also very simple:

Examples:

Find the derivative of the function.
What is the derivative of the function?

Answers: The exponential and natural logarithm are uniquely simple functions from a derivative perspective. Exponential and logarithmic functions with any other base will have a different derivative, which we will analyze later, after we go through the rules of differentiation.

Rules of differentiation

Rules of what? Again a new term, again?!...

Differentiation is the process of finding the derivative.

That's all. What else can you call this process in one word? Not derivative... The differential of mathematicians is the same increment of a function at. This term comes from the Latin differentia - difference. Here.

When deriving all these rules, we will use two functions, for example, and. We will also need formulas for their increments:

There are 5 rules in total.

The constant is taken out of the derivative sign.

If - some constant number (constant), then.

Obviously, this rule also works for the difference: .

Let's prove it. Let it be, or simpler.

Examples.

Find the derivatives of the functions:

at a point;
at a point;
at a point;
at the point.

Solutions:

(the derivative is the same at all points, since this linear function, remember?);

Derivative of the product

Everything is similar here: let’s introduce a new function and find its increment:

Derivative:

Examples:

Find the derivatives of the functions and;
Find the derivative of the function at a point.

Solutions:

Derivative of an exponential function

Now your knowledge is enough to learn how to find the derivative of any exponential function, and not just exponents (have you forgotten what that is yet?).

So, where is some number.

We already know the derivative of the function, so let's try to bring our function to a new base:

For this we will use simple rule: . Then:

Well, it worked. Now try to find the derivative, and don't forget that this function is complex.

Happened?

Here, check yourself:

The formula turned out to be very similar to the derivative of an exponent: as it was, it remains the same, only a factor appeared, which is just a number, but not a variable.

Examples:
Find the derivatives of the functions:

Answers:

This is just a number that cannot be calculated without a calculator, that is, it cannot be written down in any more in simple form. Therefore, we leave it in this form in the answer.

Note that here is a quotient of two functions, so we apply the corresponding differentiation rule:

In this example, the product of two functions:

Derivative of a logarithmic function

It’s similar here: you already know the derivative of the natural logarithm:

Therefore, to find an arbitrary logarithm with a different base, for example:

We need to reduce this logarithm to the base. How do you change the base of a logarithm? I hope you remember this formula:

Only now we will write instead:

The denominator is simply a constant (a constant number, without a variable). The derivative is obtained very simply:

Derivatives of exponential and logarithmic functions are almost never found in the Unified State Examination, but it will not be superfluous to know them.

Derivative of a complex function.

What is a "complex function"? No, this is not a logarithm, and not an arctangent. These functions can be difficult to understand (although if you find the logarithm difficult, read the topic “Logarithms” and you will be fine), but from a mathematical point of view, the word “complex” does not mean “difficult”.

Imagine a small conveyor belt: two people are sitting and doing some actions with some objects. For example, the first one wraps a chocolate bar in a wrapper, and the second one ties it with a ribbon. The result is a composite object: a chocolate bar wrapped and tied with a ribbon. To eat a chocolate bar, you need to do the reverse steps in reverse order.

Let's create a similar mathematical pipeline: first we will find the cosine of a number, and then square the resulting number. So, we are given a number (chocolate), I find its cosine (wrapper), and then you square what I got (tie it with a ribbon). What happened? Function. This is an example complex function: when, to find its value, we perform the first action directly with the variable, and then a second action with what resulted from the first.

In other words, a complex function is a function whose argument is another function: .

For our example, .

We can easily do the same steps in reverse order: first you square it, and I then look for the cosine of the resulting number: . It’s easy to guess that the result will almost always be different. An important feature of complex functions: when the order of actions changes, the function changes.

Second example: (same thing). .

The action we do last will be called "external" function, and the action performed first - accordingly "internal" function(these are informal names, I use them only to explain the material in simple language).

Try to determine for yourself which function is external and which internal:

Answers: Separating inner and outer functions is very similar to changing variables: for example, in a function

What action will we perform first? First, let's calculate the sine, and only then cube it. This means that it is an internal function, but an external one.
And the original function is their composition: .
Internal: ; external: .
Examination: .
Internal: ; external: .
Examination: .
Internal: ; external: .
Examination: .
Internal: ; external: .
Examination: .

We change variables and get a function.

Well, now we will extract our chocolate bar and look for the derivative. The procedure is always reversed: first we look for the derivative of the outer function, then we multiply the result by the derivative of the inner function. In relation to the original example, it looks like this:

Another example:

So, let's finally formulate the official rule:

Algorithm for finding the derivative of a complex function:

It seems simple, right?

Let's check with examples:

Solutions:

1) Internal: ;

External: ;

2) Internal: ;

(just don’t try to cut it by now! Nothing comes out from under the cosine, remember?)

3) Internal: ;

External: ;

It is immediately clear that this is a three-level complex function: after all, this is already a complex function in itself, and we also extract the root from it, that is, we perform the third action (put the chocolate in a wrapper and with a ribbon in the briefcase). But there is no reason to be afraid: we will still “unpack” this function in the same order as usual: from the end.

That is, first we differentiate the root, then the cosine, and only then the expression in brackets. And then we multiply it all.

In such cases, it is convenient to number the actions. That is, let's imagine what we know. In what order will we perform actions to calculate the value of this expression? Let's look at an example:

The later the action is performed, the more “external” the corresponding function will be. The sequence of actions is the same as before:

Here the nesting is generally 4-level. Let's determine the course of action.

1. Radical expression. .

2. Root. .

3. Sine. .

4. Square. .

5. Putting it all together:

DERIVATIVE. BRIEFLY ABOUT THE MAIN THINGS

Derivative of a function- the ratio of the increment of the function to the increment of the argument for an infinitesimal increment of the argument:

Basic derivatives:

Rules of differentiation:

The constant is taken out of the derivative sign:

Derivative of the sum:

Derivative of the product:

Derivative of the quotient:

Derivative of a complex function:

Algorithm for finding the derivative of a complex function:

We define the “internal” function and find its derivative.
We define the “external” function and find its derivative.
We multiply the results of the first and second points.

We present a summary table for convenience and clarity when studying the topic.

Constanty = C

Power function y = x p

(x p) " = p x p - 1

Exponential functiony = a x

(a x) " = a x ln a

In particular, whena = ewe have y = e x

(e x) " = e x

Logarithmic function

(log a x) " = 1 x ln a

In particular, whena = ewe have y = logx

(ln x) " = 1 x

Trigonometric functions

(sin x) " = cos x (cos x) " = - sin x (t g x) " = 1 cos 2 x (c t g x) " = - 1 sin 2 x

Inverse trigonometric functions

(a r c sin x) " = 1 1 - x 2 (a r c cos x) " = - 1 1 - x 2 (a r c t g x) " = 1 1 + x 2 (a r c c t g x) " = - 1 1 + x 2

Hyperbolic functions

(s h x) " = c h x (c h x) " = s h x (t h x) " = 1 c h 2 x (c t h x) " = - 1 s h 2 x

Let us analyze how the formulas of the specified table were obtained or, in other words, we will prove the derivation of derivative formulas for each type of function.

Derivative of a constant

Evidence 1

In order to derive this formula, we take as a basis the definition of the derivative of a function at a point. We use x 0 = x, where x takes the value of any real number, or, in other words, x is any number from the domain of the function f (x) = C. Let's write down the limit of the ratio of the increment of a function to the increment of the argument as ∆ x → 0:

lim ∆ x → 0 ∆ f (x) ∆ x = lim ∆ x → 0 C - C ∆ x = lim ∆ x → 0 0 ∆ x = 0

Please note that the expression 0 ∆ x falls under the limit sign. It is not the uncertainty “zero divided by zero,” since the numerator does not contain an infinitesimal value, but precisely zero. In other words, the increment of a constant function is always zero.

So, the derivative of the constant function f (x) = C is equal to zero throughout the entire domain of definition.

Example 1

The constant functions are given:

f 1 (x) = 3, f 2 (x) = a, a ∈ R, f 3 (x) = 4. 13 7 22 , f 4 (x) = 0 , f 5 (x) = - 8 7

Solution

Let us describe the given conditions. In the first function we see the derivative of the natural number 3. In the following example, you need to take the derivative of A, Where A- any real number. The third example gives us the derivative irrational number 4 . 13 7 22, the fourth is the derivative of zero (zero is an integer). Finally, in the fifth case we have the derivative rational fraction - 8 7 .

Answer: derivatives specified functions is zero for any real x(over the entire definition area)

f 1 " (x) = (3) " = 0 , f 2 " (x) = (a) " = 0 , a ∈ R , f 3 " (x) = 4 . 13 7 22 " = 0 , f 4 " (x) = 0 " = 0 , f 5 " (x) = - 8 7 " = 0

Derivative of a power function

Let's move on to the power function and the formula for its derivative, which has the form: (x p) " = p x p - 1, where the exponent p is any real number.

Evidence 2

Let us give a proof of the formula when the exponent is natural number: p = 1, 2, 3, …

We again rely on the definition of a derivative. Let's write down the limit of the ratio of the increment of a power function to the increment of the argument:

(x p) " = lim ∆ x → 0 = ∆ (x p) ∆ x = lim ∆ x → 0 (x + ∆ x) p - x p ∆ x

To simplify the expression in the numerator, we use Newton’s binomial formula:

(x + ∆ x) p - x p = C p 0 + x p + C p 1 · x p - 1 · ∆ x + C p 2 · x p - 2 · (∆ x) 2 + . . . + + C p p - 1 · x · (∆ x) p - 1 + C p p · (∆ x) p - x p = = C p 1 · x p - 1 · ∆ x + C p 2 · x p - 2 · (∆ x) 2 + . . . + C p p - 1 x (∆ x) p - 1 + C p p (∆ x) p

Thus:

(x p) " = lim ∆ x → 0 ∆ (x p) ∆ x = lim ∆ x → 0 (x + ∆ x) p - x p ∆ x = = lim ∆ x → 0 (C p 1 x p - 1 ∆ x + C p 2 · x p - 2 · (∆ x) 2 + . + C p p - 1 · x · (∆ x) p - 1 + C p p · (∆ x) p) ∆ x = = lim ∆ x → 0 (C p 1 x p - 1 + C p 2 x p - 2 ∆ x + . . + C p p - 1 x (∆ x) p - 2 + C p p (∆ x) p - 1) = = C p 1 · x p - 1 + 0 + 0 = p ! (p - 1) !

Thus, we have proven the formula for the derivative of a power function when the exponent is a natural number.

Evidence 3

To provide proof for the case when p- any real number other than zero, we use the logarithmic derivative (here we should understand the difference from the derivative logarithmic function). To have a more complete understanding, it is advisable to study the derivative of a logarithmic function and further understand the derivative of an implicit function and the derivative of a complex function.

Let's consider two cases: when x positive and when x negative.

So x > 0. Then: x p > 0 . Let us logarithm the equality y = x p to base e and apply the property of the logarithm:

y = x p ln y = ln x p ln y = p · ln x

At this stage, we have obtained an implicitly specified function. Let's define its derivative:

(ln y) " = (p · ln x) 1 y · y " = p · 1 x ⇒ y " = p · y x = p · x p x = p · x p - 1

Now we consider the case when x – a negative number.

If the indicator p is an even number, then the power function is defined for x< 0 , причем является четной: y (x) = - y ((- x) p) " = - p · (- x) p - 1 · (- x) " = = p · (- x) p - 1 = p · x p - 1

Then x p< 0 и возможно составить доказательство, используя логарифмическую производную.

If p There is odd number, then the power function is defined for x< 0 , причем является нечетной: y (x) = - y (- x) = - (- x) p . Тогда x p < 0 , а значит логарифмическую производную задействовать нельзя. В такой ситуации возможно взять за основу доказательства правила дифференцирования и правило нахождения производной сложной функции:

y " (x) = (- (- x) p) " = - ((- x) p) " = - p · (- x) p - 1 · (- x) " = = p · (- x) p - 1 = p x p - 1

The last transition is possible due to the fact that if p is an odd number, then p - 1 either an even number or zero (for p = 1), therefore, for negative x the equality (- x) p - 1 = x p - 1 is true.

So, we have proven the formula for the derivative of a power function for any real p.

Example 2

Functions given:

f 1 (x) = 1 x 2 3 , f 2 (x) = x 2 - 1 4 , f 3 (x) = 1 x log 7 12

Determine their derivatives.

Solution

We transform some of the given functions into tabular form y = x p , based on the properties of the degree, and then use the formula:

f 1 (x) = 1 x 2 3 = x - 2 3 ⇒ f 1 " (x) = - 2 3 x - 2 3 - 1 = - 2 3 x - 5 3 f 2 " (x) = x 2 - 1 4 = 2 - 1 4 x 2 - 1 4 - 1 = 2 - 1 4 x 2 - 5 4 f 3 (x) = 1 x log 7 12 = x - log 7 12 ⇒ f 3" ( x) = - log 7 12 x - log 7 12 - 1 = - log 7 12 x - log 7 12 - log 7 7 = - log 7 12 x - log 7 84

Derivative of an exponential function

Proof 4

Let us derive the derivative formula using the definition as a basis:

(a x) " = lim ∆ x → 0 a x + ∆ x - a x ∆ x = lim ∆ x → 0 a x (a ∆ x - 1) ∆ x = a x lim ∆ x → 0 a ∆ x - 1 ∆ x = 0 0

We got uncertainty. To expand it, let's write a new variable z = a ∆ x - 1 (z → 0 as ∆ x → 0). In this case, a ∆ x = z + 1 ⇒ ∆ x = log a (z + 1) = ln (z + 1) ln a . For the last transition, the formula for transition to a new logarithm base was used.

Let us substitute into the original limit:

(a x) " = a x · lim ∆ x → 0 a ∆ x - 1 ∆ x = a x · ln a · lim ∆ x → 0 1 1 z · ln (z + 1) = = a x · ln a · lim ∆ x → 0 1 ln (z + 1) 1 z = a x · ln a · 1 ln lim ∆ x → 0 (z + 1) 1 z

Let's remember the second wonderful limit and then we get the formula for the derivative of the exponential function:

(a x) " = a x · ln a · 1 ln lim z → 0 (z + 1) 1 z = a x · ln a · 1 ln e = a x · ln a

Example 3

The exponential functions are given:

f 1 (x) = 2 3 x , f 2 (x) = 5 3 x , f 3 (x) = 1 (e) x

It is necessary to find their derivatives.

Solution

We use the formula for the derivative of the exponential function and the properties of the logarithm:

f 1 " (x) = 2 3 x " = 2 3 x ln 2 3 = 2 3 x (ln 2 - ln 3) f 2 " (x) = 5 3 x " = 5 3 x ln 5 1 3 = 1 3 5 3 x ln 5 f 3 " (x) = 1 (e) x " = 1 e x " = 1 e x ln 1 e = 1 e x ln e - 1 = - 1 e x

Derivative of a logarithmic function

Evidence 5

Let us provide a proof of the formula for the derivative of a logarithmic function for any x in the domain of definition and any permissible values of the base a of the logarithm. Based on the definition of derivative, we get:

(log a x) " = lim ∆ x → 0 log a (x + ∆ x) - log a x ∆ x = lim ∆ x → 0 log a x + ∆ x x ∆ x = = lim ∆ x → 0 1 ∆ x log a 1 + ∆ x x = lim ∆ x → 0 log a 1 + ∆ x x 1 ∆ x = = lim ∆ x → 0 log a 1 + ∆ x x 1 ∆ x · x x = lim ∆ x → 0 1 x · log a 1 + ∆ x x x ∆ x = = 1 x · log a lim ∆ x → 0 1 + ∆ x x x ∆ x = 1 x · log a e = 1 x · ln e ln a = 1 x · ln a

From the indicated chain of equalities it is clear that the transformations were based on the property of the logarithm. The equality lim ∆ x → 0 1 + ∆ x x x ∆ x = e is true in accordance with the second remarkable limit.

Example 4

Logarithmic functions are given:

f 1 (x) = log ln 3 x , f 2 (x) = ln x

It is necessary to calculate their derivatives.

Solution

Let's apply the derived formula:

f 1 " (x) = (log ln 3 x) " = 1 x · ln (ln 3) ; f 2 " (x) = (ln x) " = 1 x ln e = 1 x

So, the derivative of the natural logarithm is one divided by x.

Derivatives of trigonometric functions

Proof 6

Let's use some trigonometric formulas and the first remarkable limit to derive the formula for the derivative of a trigonometric function.

According to the definition of the derivative of the sine function, we get:

(sin x) " = lim ∆ x → 0 sin (x + ∆ x) - sin x ∆ x

The formula for the difference of sines will allow us to perform the following actions:

(sin x) " = lim ∆ x → 0 sin (x + ∆ x) - sin x ∆ x = = lim ∆ x → 0 2 sin x + ∆ x - x 2 cos x + ∆ x + x 2 ∆ x = = lim ∆ x → 0 sin ∆ x 2 · cos x + ∆ x 2 ∆ x 2 = = cos x + 0 2 · lim ∆ x → 0 sin ∆ x 2 ∆ x 2

Finally, we use the first wonderful limit:

sin " x = cos x + 0 2 · lim ∆ x → 0 sin ∆ x 2 ∆ x 2 = cos x

So, the derivative of the function sin x will cos x.

We will also prove the formula for the derivative of the cosine:

cos " x = lim ∆ x → 0 cos (x + ∆ x) - cos x ∆ x = = lim ∆ x → 0 - 2 sin x + ∆ x - x 2 sin x + ∆ x + x 2 ∆ x = = - lim ∆ x → 0 sin ∆ x 2 sin x + ∆ x 2 ∆ x 2 = = - sin x + 0 2 lim ∆ x → 0 sin ∆ x 2 ∆ x 2 = - sin x

Those. the derivative of the cos x function will be – sin x.

We derive the formulas for the derivatives of tangent and cotangent based on the rules of differentiation:

t g " x = sin x cos x " = sin " x · cos x - sin x · cos " x cos 2 x = = cos x · cos x - sin x · (- sin x) cos 2 x = sin 2 x + cos 2 x cos 2 x = 1 cos 2 x c t g " x = cos x sin x " = cos " x · sin x - cos x · sin " x sin 2 x = = - sin x · sin x - cos x · cos x sin 2 x = - sin 2 x + cos 2 x sin 2 x = - 1 sin 2 x

Derivatives of inverse trigonometric functions

Derivative Section inverse functions provides comprehensive information on the proof of the formulas for the derivatives of arcsine, arccosine, arctangent and arccotangent, so we will not duplicate the material here.

Derivatives of hyperbolic functions

Evidence 7

We can derive the formulas for the derivatives of the hyperbolic sine, cosine, tangent and cotangent using the differentiation rule and the formula for the derivative of the exponential function:

s h " x = e x - e - x 2 " = 1 2 e x " - e - x " = = 1 2 e x - - e - x = e x + e - x 2 = c h x c h " x = e x + e - x 2 " = 1 2 e x " + e - x " = = 1 2 e x + - e - x = e x - e - x 2 = s h x t h " x = s h x c h x " = s h " x · c h x - s h x · c h " x c h 2 x = c h 2 x - s h 2 x c h 2 x = 1 c h 2 x c t h " x = c h x s h x " = c h " x · s h x - c h x · s h " x s h 2 x = s h 2 x - c h 2 x s h 2 x = - 1 s h 2 x

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Complex derivatives. Logarithmic derivative.
Derivative of a power-exponential function

We continue to improve our differentiation technique. In this lesson, we will consolidate the material we have covered, look at more complex derivatives, and also get acquainted with new techniques and tricks for finding a derivative, in particular, with the logarithmic derivative.

Those readers who have a low level of preparation should refer to the article How to find the derivative? Examples of solutions, which will allow you to improve your skills almost from scratch. Next, you need to carefully study the page Derivative of a complex function, understand and solve All the examples I gave. This lesson is logically the third, and after mastering it you will confidently differentiate fairly complex functions. It is undesirable to take the position of “Where else? Yes, that’s enough! ”, since all examples and solutions are taken from real tests and are often encountered in practice.

Let's start with repetition. At the lesson Derivative of a complex function We looked at a number of examples with detailed comments. During the study of differential calculus and other sections mathematical analysis– you will have to differentiate very often, and it is not always convenient (and not always necessary) to describe examples in great detail. Therefore, we will practice finding derivatives orally. The most suitable “candidates” for this are derivatives of the simplest of complex functions, for example:

According to the rule of differentiation of complex functions :

When studying other matan topics in the future, such a detailed record is most often not required; it is assumed that the student knows how to find such derivatives on autopilot. Let’s imagine that at 3 o’clock in the morning the phone rang and a pleasant voice asked: “What is the derivative of the tangent of two X’s?” This should be followed by an almost instantaneous and polite response: .

The first example will be immediately intended for independent decision.

Example 1

Find the following derivatives orally, in one action, for example: . To complete the task you only need to use table of derivatives of elementary functions(if you haven't remembered it yet). If you have any difficulties, I recommend re-reading the lesson Derivative of a complex function.

, , ,
, , ,
, , ,

, , ,

, ,

Answers at the end of the lesson

Complex derivatives

After preliminary artillery preparation, examples with 3-4-5 nestings of functions will be less scary. The following two examples may seem complicated to some, but if you understand them (someone will suffer), then almost everything else in differential calculus will seem like a child's joke.

Example 2

Find the derivative of a function

As already noted, when finding the derivative of a complex function, first of all, it is necessary Right UNDERSTAND your investments. In cases where there are doubts, I remind you of a useful technique: we take the experimental value of “x”, for example, and try (mentally or in a draft) to substitute given value into a "terrible expression".

1) First we need to calculate the expression, which means the sum is the deepest embedding.

2) Then you need to calculate the logarithm:

4) Then cube the cosine:

5) At the fifth step the difference:

6) And finally, the most external function is Square root:

Formula for differentiating a complex function are applied in reverse order, from the outermost function to the innermost. We decide:

There seem to be no errors...

(1) Take the derivative of the square root.

(2) We take the derivative of the difference using the rule

(3) The derivative of a triple is zero. In the second term we take the derivative of the degree (cube).

(4) Take the derivative of the cosine.

(5) Take the derivative of the logarithm.

(6) And finally, we take the derivative of the deepest embedding .

It may seem too difficult, but this is not the most brutal example. Take, for example, Kuznetsov’s collection and you will appreciate all the beauty and simplicity of the analyzed derivative. I noticed that they like to give a similar thing in an exam to check whether a student understands how to find the derivative of a complex function or does not understand.

The following example is for you to solve on your own.

Example 3

Find the derivative of a function

Hint: First we apply the linearity rules and the product differentiation rule

Complete solution and the answer at the end of the lesson.

It's time to move on to something smaller and nicer.
It is not uncommon for an example to show the product of not two, but three functions. How to find the derivative of the product of three factors?

Example 4

Find the derivative of a function

First, let’s see if it’s possible to turn the product of three functions into the product of two functions? For example, if we had two polynomials in the product, we could open the brackets. But in the example under consideration, all the functions are different: degree, exponent and logarithm.

In such cases it is necessary sequentially apply the product differentiation rule twice

The trick is that by “y” we denote the product of two functions: , and by “ve” we denote the logarithm: . Why can this be done? Is it really – this is not a product of two factors and the rule does not work?! There is nothing complicated:

Now it remains to apply the rule a second time to bracket:

You can also get twisted and take something out of brackets, but in this case it’s better to leave the answer exactly in this form - it will be easier to check.

The considered example can be solved in the second way:

Both solutions are absolutely equivalent.

Example 5

Find the derivative of a function

This is an example for an independent solution; in the sample it is solved using the first method.

Let's look at similar examples with fractions.

Example 6

Find the derivative of a function

There are several ways you can go here:

Or like this:

But the solution will be written more compactly if we first use the rule of differentiation of the quotient , taking for the entire numerator:

In principle, the example is solved, and if it is left as is, it will not be an error. But if you have time, it is always advisable to check on the draft to see if the answer can be simplified? Let us reduce the expression of the numerator to a common denominator and let's get rid of the three-story fraction:

The disadvantage of additional simplifications is that there is a risk of making a mistake not when finding the derivative, but during banal school transformations. On the other hand, teachers often reject the assignment and ask to “bring it to mind” the derivative.

A simpler example to solve on your own:

Example 7

Find the derivative of a function

We continue to master the methods of finding the derivative, and now we will consider a typical case when a “terrible” logarithm is proposed for differentiation

Example 8

Find the derivative of a function

Here you can go the long way, using the rule for differentiating a complex function:

But the very first step immediately plunges you into despondency - you have to take the unpleasant derivative from a fractional power, and then also from a fraction.

That's why before how to take the derivative of a “sophisticated” logarithm, it is first simplified using well-known school properties:

! If you have a practice notebook at hand, copy these formulas directly there. If you don't have a notebook, copy them onto a piece of paper, since the remaining examples of the lesson will revolve around these formulas.

The solution itself can be written something like this:

Let's transform the function:

Finding the derivative:

Pre-converting the function itself greatly simplified the solution. Thus, when a similar logarithm is proposed for differentiation, it is always advisable to “break it down”.

And now a couple of simple examples for you to solve on your own:

Example 9

Find the derivative of a function

Example 10

Find the derivative of a function

All transformations and answers are at the end of the lesson.

Logarithmic derivative

If the derivative of logarithms is such sweet music, then the question arises: is it possible in some cases to organize the logarithm artificially? Can! And even necessary.

Example 11

Find the derivative of a function

We recently looked at similar examples. What to do? You can sequentially apply the rule of differentiation of the quotient, and then the rule of differentiation of the product. The disadvantage of this method is that you end up with a huge three-story fraction, which you don’t want to deal with at all.

But in theory and practice there is such a wonderful thing as the logarithmic derivative. Logarithms can be organized artificially by “hanging” them on both sides:

Note : because function can accept negative values, then, generally speaking, you need to use modules: , which will disappear as a result of differentiation. However, the current design is also acceptable, where by default it is taken into account complex meanings. But if in all rigor, then in both cases a reservation should be made that.

Now you need to “break up” the logarithm of the right side as much as possible (formulas before your eyes?). I will describe this process in great detail:

Let's start with differentiation.
We conclude both parts under the prime:

The derivative of the right-hand side is quite simple; I will not comment on it, because if you are reading this text, you should be able to handle it confidently.

What about the left side?

On the left side we have complex function. I foresee the question: “Why, is there one letter “Y” under the logarithm?”

The fact is that this “one letter game” - IS ITSELF A FUNCTION(if it is not very clear, refer to the article Derivative of a function specified implicitly). Therefore, the logarithm is an external function, and the “y” is an internal function. And we use the rule for differentiating a complex function :

On the left side, as if by magic magic wand we have a derivative . Next, according to the rule of proportion, we transfer the “y” from the denominator of the left side to the top of the right side:

And now let’s remember what kind of “player”-function we talked about during differentiation? Let's look at the condition:

Final answer:

Example 12

Find the derivative of a function

This is an example for you to solve on your own. A sample design of an example of this type is at the end of the lesson.

Using the logarithmic derivative it was possible to solve any of examples No. 4-7, another thing is that the functions there are simpler, and perhaps the use of the logarithmic derivative is not very justified.

Derivative of a power-exponential function

We have not considered this function yet. A power-exponential function is a function for which both the degree and the base depend on the “x”. A classic example that will be given to you in any textbook or lecture:

How to find the derivative of a power-exponential function?

It is necessary to use the technique just discussed - the logarithmic derivative. We hang logarithms on both sides:

As a rule, on the right side the degree is taken out from under the logarithm:

As a result, on the right side we have the product of two functions, which will be differentiated according to the standard formula .

We find the derivative; to do this, we enclose both parts under strokes:

Further actions are simple:

Finally:

If any conversion is not entirely clear, please re-read the explanations of Example No. 11 carefully.

IN practical tasks The power-exponential function will always be more complex than the example discussed in the lecture.

Example 13

Find the derivative of a function

We use the logarithmic derivative.

On the right side we have a constant and the product of two factors - “x” and “logarithm of logarithm x” (another logarithm is nested under the logarithm). When differentiating, as we remember, it is better to immediately move the constant out of the derivative sign so that it does not get in the way; and, of course, we apply the familiar rule :