Differentiation is the calculation of the derivative.
1. Differentiation formulas.
The main differentiation formulas are in the table. They don't have to be memorized. Having understood some patterns, you will be able to independently derive others from some formulas.
1) Let's start with the formula (k x+ m)′ = k.
Its special cases are the formulas x′ = 1 and C′ = 0.
In any function of the form y = kx + m, the derivative is equal to slope k.
For example, given the function y = 2 X+ 4. Its derivative at any point will be equal to 2:
(2 x + 4)′ = 2 .
Derivative of a function at = 9 X+ 5 at any point is equal to 9 . Etc.
Let's find the derivative of the function y = 5 X. To do this, let's imagine 5 X in the form (5 X+ 0). We received an expression similar to the previous one. Means:
(5X)′ = (5 X+ 0)′ = 5.
Finally, let's find out what it is equal to x′.
Let's apply the technique from the previous example: imagine X as 1 X+ 0. Then we get:
x′ = (1 X+ 0)′ = 1.
Thus, we independently derived the formula from the table:
(0 · x+ m)′ = 0.
But then it turns out that m′ is also equal to 0. Let m = C, where C is an arbitrary constant. Then we come to another truth: the derivative of a constant is equal to zero. That is, we get another formula from the table.
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Derivatives table elementary functions
Definition 1
The derivative calculation is called differentiation.
Denote the derivative $y"$ or $\frac(dy)(dx)$.
Note 1
To find the derivative of a function, according to the basic rules of differentiation, it is transformed into another function.
Let's look at the table of derivatives. Let us pay attention to the fact that functions, after finding their derivatives, are transformed into other functions.
The only exception is $y=e^x$, which turns into itself.
Rules for differentiation of derivatives
Most often, when finding a derivative, you need to not just look at the table of derivatives, but first apply the rules of differentiation and proof of the derivative of a product, and only then use the table of derivatives of elementary functions.
1. The constant is taken out of the derivative sign
$C$ is a constant.
Example 1
Differentiate the function $y=7x^4$.
Solution.
Find $y"=(7x^4)"$. Taking the number $7$ out of the derivative sign, we get:
$y"=(7x^4)"=7(x^4)"=$
Using the table, you need to find the value of the derivative of the power function:
$=7 \cdot 4x^3=$
Let's transform the result to the form accepted in mathematics:
Answer:$28x^3$.
2. The derivative of the sum (difference) is equal to the sum (difference) of the derivatives:
$(u \pm v)"=u" \pm v"$.
Example 2
Differentiate the function $y=7+x-5x^3+4 \sin x-9\sqrt(x^2)+\frac(4)(x^4) -11\cot x$.
Solution.
$y"=(7+x-5x^5+4 \sin x-9\sqrt(x^2)+\frac(4)(x^4) -11\cot x)"=$
We apply the rule for differentiating the derivative sum and difference:
$=(7)"+(x)"-(5x^5)"+(4 \sin x)"-(9\sqrt(x^2))"+(\frac(4)(x^4) )"-(11\cot x)"=$
note that when differentiating, all powers and roots must be transformed to the form $x^(\frac(a)(b))$;
Let's take all constants out of the derivative sign:
$=(7)"+(x)"-(5x^5)"+(4\sin x)"-(9x^(\frac(2)(5)))"+(4x^(-4) )"-(11\cot x)"=$
$=(7)"+(x)"-5(x^5)"+4(\sin x)"-9(x^(\frac(2)(5)))"+4(x^( -4))"-11(\cot x)"=$
Having understood the rules of differentiation, some of them (for example, like the last two) are applied simultaneously to avoid rewriting a long expression;
we have obtained an expression from elementary functions under the derivative sign; Let's use the table of derivatives:
$=0+1-5 \cdot 5x^4+4\cos x-9 \cdot \frac(2)(5) x^(-\frac(3)(5))+12x^(-5)- 11 \cdot \frac(-1)(\sin^2 x)=$
Let's transform it to the form accepted in mathematics:
$=1-25x^4+4 \cos x-\frac(18)(5\sqrt(x^3))+\frac(12)(x^5) +\frac(11)(\sin^2 x)$
Please note that when finding the result, the terms with fractional powers convert to roots, and with negative ones - to fractions.
Answer: $1-25x^4+4 \cos x-\frac(18)(5\sqrt(x^3))+\frac(12)(x^5) +\frac(11)(\sin^2 x )$.
3. Formula for the derivative of the product of functions:
$(uv)"=u" v+uv"$.
Example 3
Differentiate the function $y=x^(11) \ln x$.
Solution.
First, we apply the rule for calculating the derivative of a product of functions, and then we use the table of derivatives:
$y"=(x^(11) \ln x)"=(x^(11))" \ln x+x^(11) (\lnтx)"=11x^(10) \ln x+x^ (11) \cdot \frac(1)(x)=11x^(10) \ln x-\frac(x^(11))(x)=11x^(10) \ln x-x^(10)=x ^(10) (11 \ln x-1)$.
Answer: $x^(10) (11 \ln x-1)$.
4. Formula for the derivative of a partial function:
$(\frac(u)(v))"=\frac(u" v-uv")(v^2)$.
Example 4
Differentiate the function $y=\frac(3x-8)(x^5-7)$.
Solution.
$y"=(\frac(3x-8)(x^5-7))"=$
according to priority rules mathematical operations First we perform division, and then addition and subtraction, so we first apply the rule for calculating the derivative of a quotient:
$=\frac((3x-8)" (x^5-7)-(3x-8) (x^5-7)")((x^5-7)^2) =$
Let’s apply the rules of sum and difference derivatives, open the brackets and simplify the expression:
$=\frac(3(x^5-7)-5x^4 (3x-8))((x^5-7)^2) =\frac(3x^5-21-15x^5+40x^ 4)((x^5-7)^2) =\frac(-12x^5+40x^4-21)((x^5-7)^2)$ .
Answer:$\frac(-12x^5+40x^4-21)((x^5-7)^2)$.
Example 5
Let's differentiate the function $y=\frac(x^7-2x+3)(x)$.
Solution.
The function y is the quotient of two functions, so you can apply the rule for calculating the derivative of the quotient, but in this case you will get a cumbersome function. To simplify this function, you can divide the numerator by the denominator term by term:
$y=\frac(x^7-13x+9)(x)=x^6-13+\frac(9)(x)$.
Let us apply the rule for differentiating the sum and difference of functions to a simplified function:
$y"=(x^6-13+\frac(9)(x))"=(x^6)"+(-13)"+9(x^(-1))"=6x^5+ 0+9 \cdot (-x^(-2))=$
$=6x^5-\frac(9)(x^2)$.
Answer: $6x^5-\frac(9)(x^2)$.