Examples are given of calculating derivatives using the formula for the derivative of a complex function.
Here we give examples of calculating derivatives of the following functions:
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If a function can be represented as a complex function in the following form:
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then its derivative is determined by the formula:
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In the examples below, we will write this formula as follows:
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Where .
Here, the subscripts or , located under the derivative sign, denote the variables by which differentiation is performed.
Usually, in tables of derivatives, derivatives of functions from the variable x are given. However, x is a formal parameter. The variable x can be replaced by any other variable. Therefore, when differentiating a function from a variable, we simply change, in the table of derivatives, the variable x to the variable u.
Simple examples
Example 1
Find the derivative of a complex function
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Solution
Let's write the given function in equivalent form:
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In the table of derivatives we find:
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According to the formula for the derivative of a complex function, we have:
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Here .
Answer
Example 2
Find the derivative
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Solution
We take the constant 5 out of the derivative sign and from the table of derivatives we find:
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Here .
Answer
Example 3
Find the derivative
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Solution
We take out a constant -1
for the sign of the derivative and from the table of derivatives we find:
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From the table of derivatives we find:
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We apply the formula for the derivative of a complex function:
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Here .
Answer
More complex examples
In more complex examples, we apply the rule for differentiating a complex function several times. In this case, we calculate the derivative from the end. That is, we break the function into its component parts and find the derivatives of the simplest parts using table of derivatives. We also use rules for differentiating sums, products and fractions. Then we make substitutions and apply the formula for the derivative of a complex function.
Example 4
Find the derivative
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Solution
Let's select the simplest part of the formula and find its derivative. .
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Here we have used the notation
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We find the derivative of the next part of the original function using the results obtained. We apply the rule for differentiating the sum:
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Once again we apply the rule of differentiation of complex functions.
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Here .
Answer
Example 5
Find the derivative of the function
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Solution
Let's select the simplest part of the formula and find its derivative from the table of derivatives. .
We apply the rule of differentiation of complex functions.
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Here
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And the theorem on the derivative of a complex function, the formulation of which is as follows:
Let 1) the function $u=\varphi (x)$ have at some point $x_0$ the derivative $u_(x)"=\varphi"(x_0)$, 2) the function $y=f(u)$ have at the corresponding at the point $u_0=\varphi (x_0)$ the derivative $y_(u)"=f"(u)$. Then the complex function $y=f\left(\varphi (x) \right)$ at the mentioned point will also have a derivative equal to the product of the derivatives of the functions $f(u)$ and $\varphi (x)$:
$$ \left(f(\varphi (x))\right)"=f_(u)"\left(\varphi (x_0) \right)\cdot \varphi"(x_0) $$
or, in shorter notation: $y_(x)"=y_(u)"\cdot u_(x)"$.
In the examples in this section, all functions have the form $y=f(x)$ (i.e., we consider only functions of one variable $x$). Accordingly, in all examples the derivative $y"$ is taken with respect to the variable $x$. To emphasize that the derivative is taken with respect to the variable $x$, $y"_x$ is often written instead of $y"$.
Examples No. 1, No. 2 and No. 3 outline the detailed process for finding the derivative of complex functions. Example No. 4 is intended for a more complete understanding of the derivative table and it makes sense to familiarize yourself with it.
It is advisable, after studying the material in examples No. 1-3, to move on to independently solving examples No. 5, No. 6 and No. 7. Examples #5, #6 and #7 contain a short solution so that the reader can check the correctness of his result.
Example No. 1
Find the derivative of the function $y=e^(\cos x)$.
We need to find the derivative of a complex function $y"$. Since $y=e^(\cos x)$, then $y"=\left(e^(\cos x)\right)"$. To find the derivative $ \left(e^(\cos x)\right)"$ we use formula No. 6 from derivative tables. In order to use formula No. 6, we need to take into account that in our case $u=\cos x$. The further solution consists in simply substituting the expression $\cos x$ instead of $u$ into formula No. 6:
$$ y"=\left(e^(\cos x) \right)"=e^(\cos x)\cdot (\cos x)" \tag (1.1)$$
Now we need to find the value of the expression $(\cos x)"$. We turn again to the table of derivatives, choosing formula No. 10 from it. Substituting $u=x$ into formula No. 10, we have: $(\cos x)"=-\ sin x\cdot x"$. Now let's continue equality (1.1), supplementing it with the result found:
$$ y"=\left(e^(\cos x) \right)"=e^(\cos x)\cdot (\cos x)"= e^(\cos x)\cdot (-\sin x \cdot x") \tag (1.2) $$
Since $x"=1$, we continue equality (1.2):
$$ y"=\left(e^(\cos x) \right)"=e^(\cos x)\cdot (\cos x)"= e^(\cos x)\cdot (-\sin x \cdot x")=e^(\cos x)\cdot (-\sin x\cdot 1)=-\sin x\cdot e^(\cos x) \tag (1.3) $$
So, from equality (1.3) we have: $y"=-\sin x\cdot e^(\cos x)$. Naturally, explanations and intermediate equalities are usually skipped, writing down the finding of the derivative in one line, as in the equality ( 1.3). So, the derivative of the complex function has been found, all that remains is to write down the answer.
Answer: $y"=-\sin x\cdot e^(\cos x)$.
Example No. 2
Find the derivative of the function $y=9\cdot \arctg^(12)(4\cdot \ln x)$.
We need to calculate the derivative $y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"$. To begin with, we note that the constant (i.e. the number 9) can be taken out of the derivative sign:
$$ y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"=9\cdot\left(\arctg^(12)(4\cdot \ln x) \right)" \tag (2.1) $$
Now let's turn to the expression $\left(\arctg^(12)(4\cdot \ln x) \right)"$. To select the desired formula from derivative tables it was easier, I will present the expression in question in this form: $\left(\left(\arctg(4\cdot \ln x) \right)^(12)\right)"$. Now it is clear that it is necessary to use formula No. 2 , i.e. $\left(u^\alpha \right)"=\alpha\cdot u^(\alpha-1)\cdot u"$. We substitute $u=\arctg(4\cdot \) into this formula ln x)$ and $\alpha=12$:
Supplementing equality (2.1) with the result obtained, we have:
$$ y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"=9\cdot\left(\arctg^(12)(4\cdot \ln x) \right)"= 108\cdot\left(\arctg(4\cdot \ln x) \right)^(11)\cdot (\arctg(4\cdot \ln x))" \tag (2.2) $$
In this situation, a mistake is often made when the solver at the first step chooses the formula $(\arctg \; u)"=\frac(1)(1+u^2)\cdot u"$ instead of the formula $\left(u^\ alpha \right)"=\alpha\cdot u^(\alpha-1)\cdot u"$. The point is that the derivative of the external function must come first. To understand which function will be external to the expression $\arctg^(12)(4\cdot 5^x)$, imagine that you are calculating the value of the expression $\arctg^(12)(4\cdot 5^x)$ at some value $x$. First you will calculate the value of $5^x$, then multiply the result by 4, getting $4\cdot 5^x$. Now we take the arctangent from this result, obtaining $\arctg(4\cdot 5^x)$. Then we raise the resulting number to the twelfth power, getting $\arctg^(12)(4\cdot 5^x)$. The last action, i.e. raising to the power of 12 will be an external function. And it is from this that we must begin to find the derivative, which was done in equality (2.2).
Now we need to find $(\arctg(4\cdot \ln x))"$. We use formula No. 19 of the derivatives table, substituting $u=4\cdot \ln x$ into it:
$$ (\arctg(4\cdot \ln x))"=\frac(1)(1+(4\cdot \ln x)^2)\cdot (4\cdot \ln x)" $$
Let's simplify the resulting expression a little, taking into account $(4\cdot \ln x)^2=4^2\cdot (\ln x)^2=16\cdot \ln^2 x$.
$$ (\arctg(4\cdot \ln x))"=\frac(1)(1+(4\cdot \ln x)^2)\cdot (4\cdot \ln x)"=\frac( 1)(1+16\cdot \ln^2 x)\cdot (4\cdot \ln x)" $$
Equality (2.2) will now become:
$$ y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"=9\cdot\left(\arctg^(12)(4\cdot \ln x) \right)"=\\ =108\cdot\left(\arctg(4\cdot \ln x) \right)^(11)\cdot (\arctg(4\cdot \ln x))"=108\cdot \left(\arctg(4\cdot \ln x) \right)^(11)\cdot \frac(1)(1+16\cdot \ln^2 x)\cdot (4\cdot \ln x)" \tag (2.3) $$
It remains to find $(4\cdot \ln x)"$. Let's take the constant (i.e. 4) out of the derivative sign: $(4\cdot \ln x)"=4\cdot (\ln x)"$. For In order to find $(\ln x)"$ we use formula No. 8, substituting $u=x$ into it: $(\ln x)"=\frac(1)(x)\cdot x"$. Since $x"=1$, then $(\ln x)"=\frac(1)(x)\cdot x"=\frac(1)(x)\cdot 1=\frac(1)(x )$. Substituting the obtained result into formula (2.3), we obtain:
$$ y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"=9\cdot\left(\arctg^(12)(4\cdot \ln x) \right)"=\\ =108\cdot\left(\arctg(4\cdot \ln x) \right)^(11)\cdot (\arctg(4\cdot \ln x))"=108\cdot \left(\arctg(4\cdot \ln x) \right)^(11)\cdot \frac(1)(1+16\cdot \ln^2 x)\cdot (4\cdot \ln x)" =\\ =108\cdot \left(\arctg(4\cdot \ln x) \right)^(11)\cdot \frac(1)(1+16\cdot \ln^2 x)\cdot 4\ cdot \frac(1)(x)=432\cdot \frac(\arctg^(11)(4\cdot \ln x))(x\cdot (1+16\cdot \ln^2 x)). $
Let me remind you that the derivative of a complex function is most often found in one line, as written in the last equality. Therefore, when preparing standard calculations or control work, it is not at all necessary to describe the solution in such detail.
Answer: $y"=432\cdot \frac(\arctg^(11)(4\cdot \ln x))(x\cdot (1+16\cdot \ln^2 x))$.
Example No. 3
Find $y"$ of the function $y=\sqrt(\sin^3(5\cdot9^x))$.
First, let's slightly transform the function $y$, expressing the radical (root) as a power: $y=\sqrt(\sin^3(5\cdot9^x))=\left(\sin(5\cdot 9^x) \right)^(\frac(3)(7))$. Now let's start finding the derivative. Since $y=\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))$, then:
$$ y"=\left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)" \tag (3.1) $$
We use formula No. 2 from derivative tables, substituting $u=\sin(5\cdot 9^x)$ and $\alpha=\frac(3)(7)$ into it:
$$ \left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)"= \frac(3)(7)\cdot \left( \sin(5\cdot 9^x)\right)^(\frac(3)(7)-1) (\sin(5\cdot 9^x))"=\frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) (\sin(5\cdot 9^x))" $$
Let us continue equality (3.1) using the result obtained:
$$ y"=\left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)"=\frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) (\sin(5\cdot 9^x))" \tag (3.2) $$
Now we need to find $(\sin(5\cdot 9^x))"$. For this we use formula No. 9 from the table of derivatives, substituting $u=5\cdot 9^x$ into it:
$$ (\sin(5\cdot 9^x))"=\cos(5\cdot 9^x)\cdot(5\cdot 9^x)" $$
Having supplemented equality (3.2) with the result obtained, we have:
$$ y"=\left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)"=\frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) (\sin(5\cdot 9^x))"=\\ =\frac(3) (7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) \cos(5\cdot 9^x)\cdot(5\cdot 9 ^x)" \tag (3.3) $$
It remains to find $(5\cdot 9^x)"$. First, let's take the constant (the number $5$) outside the derivative sign, i.e. $(5\cdot 9^x)"=5\cdot (9^x) "$. To find the derivative $(9^x)"$, apply formula No. 5 of the table of derivatives, substituting $a=9$ and $u=x$ into it: $(9^x)"=9^x\cdot \ ln9\cdot x"$. Since $x"=1$, then $(9^x)"=9^x\cdot \ln9\cdot x"=9^x\cdot \ln9$. Now we can continue equality (3.3):
$$ y"=\left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)"=\frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) (\sin(5\cdot 9^x))"=\\ =\frac(3) (7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) \cos(5\cdot 9^x)\cdot(5\cdot 9 ^x)"= \frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) \cos(5\cdot 9 ^x)\cdot 5\cdot 9^x\cdot \ln9=\\ =\frac(15\cdot \ln 9)(7)\cdot \left(\sin(5\cdot 9^x)\right) ^(-\frac(4)(7))\cdot \cos(5\cdot 9^x)\cdot 9^x. $$
We can again return from powers to radicals (i.e., roots), writing $\left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7))$ in the form $\ frac(1)(\left(\sin(5\cdot 9^x)\right)^(\frac(4)(7)))=\frac(1)(\sqrt(\sin^4(5\ cdot 9^x)))$. Then the derivative will be written in this form:
$$ y"=\frac(15\cdot \ln 9)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7))\cdot \cos(5\cdot 9^x)\cdot 9^x= \frac(15\cdot \ln 9)(7)\cdot \frac(\cos (5\cdot 9^x)\cdot 9^x) (\sqrt(\sin^4(5\cdot 9^x))).
Answer: $y"=\frac(15\cdot \ln 9)(7)\cdot \frac(\cos (5\cdot 9^x)\cdot 9^x)(\sqrt(\sin^4(5\ cdot 9^x)))$.
Example No. 4
Show that formulas No. 3 and No. 4 of the table of derivatives are a special case of formula No. 2 of this table.
Formula No. 2 of the table of derivatives contains the derivative of the function $u^\alpha$. Substituting $\alpha=-1$ into formula No. 2, we get:
$$(u^(-1))"=-1\cdot u^(-1-1)\cdot u"=-u^(-2)\cdot u"\tag (4.1)$$
Since $u^(-1)=\frac(1)(u)$ and $u^(-2)=\frac(1)(u^2)$, then equality (4.1) can be rewritten as follows: $ \left(\frac(1)(u) \right)"=-\frac(1)(u^2)\cdot u"$. This is formula No. 3 of the table of derivatives.
Let us turn again to formula No. 2 of the table of derivatives. Let's substitute $\alpha=\frac(1)(2)$ into it:
$$\left(u^(\frac(1)(2))\right)"=\frac(1)(2)\cdot u^(\frac(1)(2)-1)\cdot u" =\frac(1)(2)u^(-\frac(1)(2))\cdot u"\tag (4.2) $$
Since $u^(\frac(1)(2))=\sqrt(u)$ and $u^(-\frac(1)(2))=\frac(1)(u^(\frac( 1)(2)))=\frac(1)(\sqrt(u))$, then equality (4.2) can be rewritten as follows:
$$ (\sqrt(u))"=\frac(1)(2)\cdot \frac(1)(\sqrt(u))\cdot u"=\frac(1)(2\sqrt(u) )\cdot u" $$
The resulting equality $(\sqrt(u))"=\frac(1)(2\sqrt(u))\cdot u"$ is formula No. 4 of the table of derivatives. As you can see, formulas No. 3 and No. 4 of the derivative table are obtained from formula No. 2 by substituting the corresponding $\alpha$ value.