Derivative of e 2x. Geometric and physical meaning of the derivative. Power function derivative
If we follow the definition, then the derivative of a function at a point is the limit of the increment ratio of the function Δ y to the increment of the argument Δ x:
Everything seems to be clear. But try to calculate by this formula, say, the derivative of the function f(x) = x 2 + (2x+ 3) · e x sin x. If you do everything by definition, then after a couple of pages of calculations you will simply fall asleep. Therefore, there are simpler and more effective ways.
To begin with, we note that the so-called elementary functions can be distinguished from the whole variety of functions. These are relatively simple expressions, the derivatives of which have long been calculated and entered in the table. Such functions are easy enough to remember, along with their derivatives.
Derivatives of elementary functions
Elementary functions are everything listed below. The derivatives of these functions must be known by heart. Moreover, it is not difficult to memorize them - that's why they are elementary.
So, the derivatives of elementary functions:
Name | Function | Derivative |
Constant | f(x) = C, C ∈ R | 0 (yes, yes, zero!) |
Degree with rational exponent | f(x) = x n | n · x n − 1 |
Sinus | f(x) = sin x | cos x |
Cosine | f(x) = cos x | − sin x(minus sine) |
Tangent | f(x) = tg x | 1/cos 2 x |
Cotangent | f(x) = ctg x | − 1/sin2 x |
natural logarithm | f(x) = log x | 1/x |
Arbitrary logarithm | f(x) = log a x | 1/(x ln a) |
Exponential function | f(x) = e x | e x(nothing changed) |
If an elementary function is multiplied by an arbitrary constant, then the derivative of the new function is also easily calculated:
(C · f)’ = C · f ’.
In general, constants can be taken out of the sign of the derivative. For example:
(2x 3)' = 2 ( x 3)' = 2 3 x 2 = 6x 2 .
Obviously, elementary functions can be added to each other, multiplied, divided, and much more. This is how new functions will appear, no longer very elementary, but also differentiable according to certain rules. These rules are discussed below.
Derivative of sum and difference
Let the functions f(x) And g(x), whose derivatives are known to us. For example, you can take the elementary functions discussed above. Then you can find the derivative of the sum and difference of these functions:
- (f + g)’ = f ’ + g ’
- (f − g)’ = f ’ − g ’
So, the derivative of the sum (difference) of two functions is equal to the sum (difference) of the derivatives. There may be more terms. For example, ( f + g + h)’ = f ’ + g ’ + h ’.
Strictly speaking, there is no concept of "subtraction" in algebra. There is a concept of "negative element". Therefore, the difference f − g can be rewritten as a sum f+ (−1) g, and then only one formula remains - the derivative of the sum.
f(x) = x 2 + sinx; g(x) = x 4 + 2x 2 − 3.
Function f(x) is the sum of two elementary functions, so:
f ’(x) = (x 2+ sin x)’ = (x 2)' + (sin x)’ = 2x+ cosx;
We argue similarly for the function g(x). Only there are already three terms (from the point of view of algebra):
g ’(x) = (x 4 + 2x 2 − 3)’ = (x 4 + 2x 2 + (−3))’ = (x 4)’ + (2x 2)’ + (−3)’ = 4x 3 + 4x + 0 = 4x · ( x 2 + 1).
Answer:
f ’(x) = 2x+ cosx;
g ’(x) = 4x · ( x
2 + 1).
Derivative of a product
Mathematics is a logical science, so many people believe that if the derivative of the sum is equal to the sum of the derivatives, then the derivative of the product strike"\u003e equal to the product of derivatives. But figs to you! The derivative of the product is calculated using a completely different formula. Namely:
(f · g) ’ = f ’ · g + f · g ’
The formula is simple, but often forgotten. And not only schoolchildren, but also students. The result is incorrectly solved problems.
Task. Find derivatives of functions: f(x) = x 3 cosx; g(x) = (x 2 + 7x− 7) · e x .
Function f(x) is a product of two elementary functions, so everything is simple:
f ’(x) = (x 3 cos x)’ = (x 3)' cos x + x 3 (cos x)’ = 3x 2 cos x + x 3 (−sin x) = x 2 (3cos x − x sin x)
Function g(x) the first multiplier is a little more complicated, but the general scheme does not change from this. Obviously, the first multiplier of the function g(x) is a polynomial, and its derivative is the derivative of the sum. We have:
g ’(x) = ((x 2 + 7x− 7) · e x)’ = (x 2 + 7x− 7)' · e x + (x 2 + 7x− 7) ( e x)’ = (2x+ 7) · e x + (x 2 + 7x− 7) · e x = e x(2 x + 7 + x 2 + 7x −7) = (x 2 + 9x) · e x = x(x+ 9) · e x .
Answer:
f ’(x) = x 2 (3cos x − x sin x);
g ’(x) = x(x+ 9) · e
x
.
Note that in the last step, the derivative is factorized. Formally, this is not necessary, but most derivatives are not calculated on their own, but to explore the function. This means that further the derivative will be equated to zero, its signs will be found out, and so on. For such a case, it is better to have an expression decomposed into factors.
If there are two functions f(x) And g(x), and g(x) ≠ 0 on the set of interest to us, we can define new feature h(x) = f(x)/g(x). For such a function, you can also find the derivative:
Not weak, right? Where did the minus come from? Why g 2? And like this! This is one of the most complex formulas - you can’t figure it out without a bottle. Therefore, it is better to study it on concrete examples.
Task. Find derivatives of functions:
There are elementary functions in the numerator and denominator of each fraction, so all we need is the formula for the derivative of the quotient:
By tradition, we factor the numerator into factors - this will greatly simplify the answer:
A complex function is not necessarily a formula half a kilometer long. For example, it suffices to take the function f(x) = sin x and replace the variable x, say, on x 2+ln x. It turns out f(x) = sin ( x 2+ln x) is a complex function. She also has a derivative, but it will not work to find it according to the rules discussed above.
How to be? In such cases, the replacement of a variable and the formula for the derivative of a complex function help:
f ’(x) = f ’(t) · t', If x is replaced by t(x).
As a rule, the situation with the understanding of this formula is even more sad than with the derivative of the quotient. Therefore, it is also better to explain it with specific examples, with detailed description every step.
Task. Find derivatives of functions: f(x) = e 2x + 3 ; g(x) = sin ( x 2+ln x)
Note that if in the function f(x) instead of expression 2 x+ 3 will be easy x, then we get an elementary function f(x) = e x. Therefore, we make a substitution: let 2 x + 3 = t, f(x) = f(t) = e t. We are looking for the derivative of a complex function by the formula:
f ’(x) = f ’(t) · t ’ = (e t)’ · t ’ = e t · t ’
And now - attention! Performing a reverse substitution: t = 2x+ 3. We get:
f ’(x) = e t · t ’ = e 2x+ 3 (2 x + 3)’ = e 2x+ 3 2 = 2 e 2x + 3
Now let's look at the function g(x). Obviously needs to be replaced. x 2+ln x = t. We have:
g ’(x) = g ’(t) · t' = (sin t)’ · t' = cos t · t ’
Reverse replacement: t = x 2+ln x. Then:
g ’(x) = cos( x 2+ln x) · ( x 2+ln x)' = cos ( x 2+ln x) · (2 x + 1/x).
That's all! As can be seen from the last expression, the whole problem has been reduced to calculating the derivative of the sum.
Answer:
f ’(x) = 2 e
2x + 3 ;
g ’(x) = (2x + 1/x) cos ( x 2+ln x).
Very often in my lessons, instead of the term “derivative”, I use the word “stroke”. For example, the stroke of the sum is equal to the sum of the strokes. Is that clearer? Well, that's good.
Thus, the calculation of the derivative comes down to getting rid of these very strokes according to the rules discussed above. As a final example, let's return to the derivative power with a rational exponent:
(x n)’ = n · x n − 1
Few know that in the role n may well be a fractional number. For example, the root is x 0.5 . But what if there is something tricky under the root? Again, a complex function will turn out - they like to give such constructions on control work and exams.
Task. Find the derivative of a function:
First, let's rewrite the root as a power with a rational exponent:
f(x) = (x 2 + 8x − 7) 0,5 .
Now we make a substitution: let x 2 + 8x − 7 = t. We find the derivative by the formula:
f ’(x) = f ’(t) · t ’ = (t 0.5)' t' = 0.5 t−0.5 t ’.
We make a reverse substitution: t = x 2 + 8x− 7. We have:
f ’(x) = 0.5 ( x 2 + 8x− 7) −0.5 ( x 2 + 8x− 7)' = 0.5 (2 x+ 8) ( x 2 + 8x − 7) −0,5 .
Finally, back to the roots:
The calculation of the derivative is often found in USE assignments. This page contains a list of formulas for finding derivatives.
Differentiation rules
- (k⋅f(x))′=k⋅f′(x).
- (f(x)+g(x))′=f′(x)+g′(x).
- (f(x)⋅ g(x))′=f′(x)⋅ g(x)+f(x)⋅ g′(x).
- Derivative of a complex function. If y=F(u) and u=u(x), then the function y=f(x)=F(u(x)) is called a complex function of x. Is equal to y′(x)=Fu′⋅ ux′.
- Derivative of an implicit function. The function y=f(x) is called the implicit function given by the relation F(x,y)=0 if F(x,f(x))≡0.
- Derivative of the inverse function. If g(f(x))=x, then the function g(x) is called the inverse function for the function y=f(x).
- Derivative of a parametrically given function. Let x and y be given as functions of the variable t: x=x(t), y=y(t). They say that y=y(x) parametrically given function on the interval x∈ (a;b), if on this interval the equation x=x(t) can be expressed as t=t(x) and the function y=y(t(x))=y(x) can be defined.
- Derivative of exponential function. It is found by taking the logarithm to the base of the natural logarithm.
- Table of derivatives of exponential and logarithmic functions
Derivatives of simple functions
1. The derivative of a number is zeroс´ = 0
Example:
5' = 0
Explanation:
The derivative shows the rate at which the value of the function changes when the 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 (calculation result) increases by the same amount. Thus, the rate of change of the value of the function y = x is exactly equal to the rate of change of 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, each time the function argument ( X) its value (y) grows in With once. Thus, the rate of change of the value of the function with respect 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 straight line (k).
4. Modulo derivative of a variable is equal to the quotient of this variable to its modulus
|x|"= x / |x| provided that x ≠ 0
Explanation:
Since the derivative of the variable (see formula 2) is equal to one, the derivative of the modulus differs only in that the value of the rate of change of the function changes to the opposite when crossing the origin point (try to draw a graph of the function y = |x| and see for yourself. This is exactly value and returns the expression x / |x| When x< 0 оно равно (-1), а когда x >0 - one. That is, at negative values variable x with each increase in the change of the argument, the value of the function decreases by exactly the same value, and for positive ones, on the contrary, it increases, but by exactly the same value.
5. Power derivative of a variable is equal to the product of the number of this power and the variable in 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 memorize the formula:
Take the exponent of the variable "down" as a multiplier, and then decrease the exponent itself by one. For example, for x 2 - two was ahead of x, and then the reduced power (2-1 = 1) just gave us 2x. The same thing happened for x 3 - we lower 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.Fraction derivative 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 derivatives table
(x -1)" = -1x -2 = - 1 / x 2
7. Fraction derivative with a variable of arbitrary degree in the denominator
(1/x c)" = - c / x c+1
Example:
(1 / x 2)" = - 2 / x 3
8. root derivative(derivative of variable under square root)
(√x)" = 1 / (2√x) or 1/2 x -1/2
Example:
(√x)" = (x 1/2)" so you can apply the formula from rule 5
(x 1/2)" \u003d 1/2 x -1/2 \u003d 1 / (2√x)
9. Derivative of a variable under a root of an arbitrary degree
(n √ x)" = 1 / (n n √ x n-1)
The derivative of the exponent is equal to the exponent itself (the derivative of e to the power of x is equal to e to the power of x):
(1)
(e x )′ = e x.
The derivative of an exponential function with a base of degree a is equal to the function itself, multiplied by the natural logarithm of a:
(2)
.
Derivation of the formula for the derivative of the exponent, e to the power of x
The exponent is an exponential function whose exponent base is equal to the number e, which is the following limit:
.
Here it can be either a natural or a real number. Next, we derive formula (1) for the derivative of the exponent.
Derivation of the formula for the derivative of the exponent
Consider the exponent, e to the power of x :
y = e x .
This function is defined for all . Let's find its derivative with respect to x . By definition, the derivative is the following limit:
(3)
.
Let's transform this expression to reduce it to known mathematical properties and rules. For this we need the following facts:
A) Exponent property:
(4)
;
B) Logarithm property:
(5)
;
IN) Continuity of the logarithm and the property of limits for continuous function:
(6)
.
Here, is some function that has a limit and this limit is positive.
G) The meaning of the second wonderful limit:
(7)
.
We apply these facts to our limit (3). We use property (4):
;
.
Let's make a substitution. Then ; .
Due to the continuity of the exponent,
.
Therefore, at , . As a result, we get:
.
Let's make a substitution. Then . At , . And we have:
.
We apply the property of the logarithm (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 have obtained formula (1) for the derivative of the exponent.
Derivation of the formula for the derivative of the exponential function
Now we derive the 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 us transform formula (8). To do this, we use the properties of the exponential function and the logarithm.
;
.
So, we have transformed formula (8) to the following form:
.
Higher order derivatives of e to the power of x
Now let's find derivatives of higher orders. Let's look at the exponent first:
(14)
.
(1)
.
We see that the derivative of the function (14) is equal to the function (14) itself. Differentiating (1), we obtain second and third order derivatives:
;
.
This shows that the nth order derivative is also equal to the original function:
.
Higher order derivatives of the exponential function
Now consider exponential function with base degree a :
.
We found its first order derivative:
(15)
.
Differentiating (15), we obtain second and third order derivatives:
;
.
We see that each differentiation leads to the multiplication of the original function by . Therefore, the nth derivative has the following form:
.