# 3.6 The chain rule  (Page 3/6)

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## Using the chain rule on a general cosine function

Find the derivative of $h\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}\left(g\left(x\right)\right).$

Think of $h\left(x\right)=\text{cos}\left(g\left(x\right)\right)$ as $f\left(g\left(x\right)\right)$ where $f\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x.$ Since ${f}^{\prime }\left(x\right)=\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}x.$ we have ${f}^{\prime }\left(g\left(x\right)\right)=\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(g\left(x\right)\right).$ Then we do the following calculation.

$\begin{array}{ccccc}\hfill {h}^{\prime }\left(x\right)& ={f}^{\prime }\left(g\left(x\right)\right){g}^{\prime }\left(x\right)\hfill & & & \text{Apply the chain rule.}\hfill \\ & =\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(g\left(x\right)\right){g}^{\prime }\left(x\right)\hfill & & & \text{Substitute}\phantom{\rule{0.2em}{0ex}}{f}^{\prime }\left(g\left(x\right)\right)=\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(g\left(x\right)\right).\hfill \end{array}$

Thus, the derivative of $h\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}\left(g\left(x\right)\right)$ is given by ${h}^{\prime }\left(x\right)=\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(g\left(x\right)\right){g}^{\prime }\left(x\right).$

In the following example we apply the rule that we have just derived.

## Using the chain rule on a cosine function

Find the derivative of $h\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}\left(5{x}^{2}\right).$

Let $g\left(x\right)=5{x}^{2}.$ Then ${g}^{\prime }\left(x\right)=10x.$ Using the result from the previous example,

$\begin{array}{cc}\hfill {h}^{\prime }\left(x\right)& =\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(5{x}^{2}\right)·10x\hfill \\ & =-10x\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(5{x}^{2}\right).\hfill \end{array}$

## Using the chain rule on another trigonometric function

Find the derivative of $h\left(x\right)=\text{sec}\phantom{\rule{0.1em}{0ex}}\left(4{x}^{5}+2x\right).$

Apply the chain rule to $h\left(x\right)=\text{sec}\phantom{\rule{0.1em}{0ex}}\left(g\left(x\right)\right)$ to obtain

${h}^{\prime }\left(x\right)=\text{sec}\left(g\left(x\right)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}\left(g\left(x\right)\right){g}^{\prime }\left(x\right).$

In this problem, $g\left(x\right)=4{x}^{5}+2x,$ so we have ${g}^{\prime }\left(x\right)=20{x}^{4}+2.$ Therefore, we obtain

$\begin{array}{cc}\hfill {h}^{\prime }\left(x\right)& =\text{sec}\phantom{\rule{0.1em}{0ex}}\left(4{x}^{5}+2x\right)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}\left(4{x}^{5}+2x\right)\left(20{x}^{4}+2\right)\hfill \\ & =\left(20{x}^{4}+2\right)\text{sec}\phantom{\rule{0.1em}{0ex}}\left(4{x}^{5}+2x\right)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}\left(4{x}^{5}+2x\right).\hfill \end{array}$

Find the derivative of $h\left(x\right)=\text{sin}\left(7x+2\right).$

${h}^{\prime }\left(x\right)=7\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}\left(7x+2\right)$

At this point we provide a list of derivative formulas that may be obtained by applying the chain rule in conjunction with the formulas for derivatives of trigonometric functions. Their derivations are similar to those used in [link] and [link] . For convenience, formulas are also given in Leibniz’s notation, which some students find easier to remember. (We discuss the chain rule using Leibniz’s notation at the end of this section.) It is not absolutely necessary to memorize these as separate formulas as they are all applications of the chain rule to previously learned formulas.

## Using the chain rule with trigonometric functions

For all values of $x$ for which the derivative is defined,

$\begin{array}{cccc}\frac{d}{dx}\left(\text{sin}\left(g\left(x\right)\right)\phantom{\rule{0.2em}{0ex}}=\text{cos}\phantom{\rule{0.1em}{0ex}}\left(g\left(x\right)\right)g\prime \left(x\right)\hfill & & & \frac{d}{dx}\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}u\phantom{\rule{0.2em}{0ex}}=\text{cos}\phantom{\rule{0.1em}{0ex}}u\frac{du}{dx}\hfill \\ \frac{d}{dx}\left(\text{cos}\left(g\left(x\right)\right)\phantom{\rule{0.1em}{0ex}}=\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(g\left(x\right)\right)g\prime \left(x\right)\hfill & & & \frac{d}{dx}\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}u\phantom{\rule{0.1em}{0ex}}=\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}u\frac{du}{dx}\hfill \\ \frac{d}{dx}\left(\text{tan}\left(g\left(x\right)\right)\phantom{\rule{0.22em}{0ex}}={\text{sec}}^{2}\left(g\left(x\right)\right)g\prime \left(x\right)\hfill & & & \frac{d}{dx}\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}u\phantom{\rule{0.2em}{0ex}}={\text{sec}}^{2}u\frac{du}{dx}\hfill \\ \frac{d}{dx}\left(\text{cot}\left(g\left(x\right)\right)\phantom{\rule{0.25em}{0ex}}=\text{−}{\text{csc}}^{2}\left(g\left(x\right)\right)g\prime \left(x\right)\hfill & & & \frac{d}{dx}\phantom{\rule{0.1em}{0ex}}\text{cot}\phantom{\rule{0.1em}{0ex}}u\phantom{\rule{0.25em}{0ex}}=\text{−}{\text{csc}}^{2}u\frac{du}{dx}\hfill \\ \frac{d}{dx}\left(\text{sec}\left(g\left(x\right)\right)\phantom{\rule{0.25em}{0ex}}=\text{sec}\left(g\left(x\right)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}\left(g\left(x\right)\right)g\prime \left(x\right)\hfill & & & \frac{d}{dx}\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}u\phantom{\rule{0.25em}{0ex}}=\text{sec}\phantom{\rule{0.1em}{0ex}}u\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}u\frac{du}{dx}\hfill \\ \frac{d}{dx}\left(\text{csc}\left(g\left(x\right)\right)\phantom{\rule{0.25em}{0ex}}=\text{−}\text{csc}\left(g\left(x\right)\right)\text{cot}\phantom{\rule{0.1em}{0ex}}\left(g\left(x\right)\right)g\prime \left(x\right)\hfill & & & \frac{d}{dx}\phantom{\rule{0.1em}{0ex}}\text{csc}\phantom{\rule{0.1em}{0ex}}u\phantom{\rule{0.25em}{0ex}}=\text{−}\text{csc}\phantom{\rule{0.1em}{0ex}}u\phantom{\rule{0.1em}{0ex}}\text{cot}\phantom{\rule{0.1em}{0ex}}u\frac{du}{dx}.\hfill \end{array}$

## Combining the chain rule with the product rule

Find the derivative of $h\left(x\right)={\left(2x+1\right)}^{5}{\left(3x-2\right)}^{7}.$

First apply the product rule, then apply the chain rule to each term of the product.

$\begin{array}{ccccc}\hfill {h}^{\prime }\left(x\right)& =\frac{d}{dx}\left({\left(2x+1\right)}^{5}\right)·{\left(3x-2\right)}^{7}+\frac{d}{dx}\left({\left(3x-2\right)}^{7}\right)·{\left(2x+1\right)}^{5}\hfill & & & \text{Apply the product rule.}\hfill \\ & =5{\left(2x+1\right)}^{4}·2·{\left(3x-2\right)}^{7}+7{\left(3x-2\right)}^{6}·3·{\left(2x+1\right)}^{5}\hfill & & & \text{Apply the chain rule.}\hfill \\ & =10{\left(2x+1\right)}^{4}{\left(3x-2\right)}^{7}+21{\left(3x-2\right)}^{6}{\left(2x+1\right)}^{5}\hfill & & & \text{Simplify.}\hfill \\ & ={\left(2x+1\right)}^{4}{\left(3x-2\right)}^{6}\left(10\left(3x-7\right)+21\left(2x+1\right)\right)\hfill & & & \text{Factor out}\phantom{\rule{0.2em}{0ex}}{\left(2x+1\right)}^{4}{\left(3x-2\right)}^{6}.\hfill \\ & ={\left(2x+1\right)}^{4}{\left(3x-2\right)}^{6}\left(72x-49\right)\hfill & & & \text{Simplify.}\hfill \end{array}$

Find the derivative of $h\left(x\right)=\frac{x}{{\left(2x+3\right)}^{3}}.$

${h}^{\prime }\left(x\right)=\frac{3-4x}{{\left(2x+3\right)}^{4}}$

## Composites of three or more functions

We can now combine the chain rule with other rules for differentiating functions, but when we are differentiating the composition of three or more functions, we need to apply the chain rule more than once. If we look at this situation in general terms, we can generate a formula, but we do not need to remember it, as we can simply apply the chain rule multiple times.

In general terms, first we let

$k\left(x\right)=h\left(f\left(g\left(x\right)\right)\right).$

Then, applying the chain rule once we obtain

${k}^{\prime }\left(x\right)=\frac{d}{dx}\left(h\left(f\left(g\left(x\right)\right)\right)=h\prime \left(f\left(g\left(x\right)\right)\right)·\frac{d}{dx}f\left(\left(g\left(x\right)\right)\right).$

Applying the chain rule again, we obtain

what is the power rule
how do i deal with infinity in limits?
f(x)=7x-x g(x)=5-x
Awon
5x-5
Verna
what is domain
difference btwn domain co- domain and range
Cabdalla
x
Verna
The set of inputs of a function. x goes in the function, y comes out.
Verna
where u from verna
Arfan
If you differentiate then answer is not x
Raymond
domain is the set of values of independent variable and the range is the corresponding set of values of dependent variable
Champro
what is functions
give different types of functions.
Paul
how would u find slope of tangent line to its inverse function, if the equation is x^5+3x^3-4x-8 at the point(-8,1)
pls solve it i Want to see the answer
Sodiq
ok
Friendz
differentiate each term
Friendz
why do we need to study functions?
to understand how to model one variable as a direct relationship to another variable
Andrew
integrate the root of 1+x²
use the substitution t=1+x. dt=dx √(1+x)dx = √tdt = t^1/2 dt integral is then = t^(1/2 + 1) / (1/2 + 1) + C = (2/3) t^(3/2) + C substitute back t=1+x = (2/3) (1+x)^(3/2) + C
navin
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f(x) = x^3 + 4, to find inverse switch x and you and isolate y: x = y^3 + 4 x -4 = y^3 (x-4)^1/3 = y = f^-1(x)
Andrew
in the example exercise how does it go from -4 +- squareroot(8)/-4 to -4 +- 2squareroot(2)/-4 what is the process of pulling out the factor like that?
Andrew
√(8) =√(4x2) =√4 x √2 2 √2 hope this helps. from the surds theory a^c x b^c = (ab)^c
Barnabas
564356
Myong
can you determine whether f(x)=x(cube) +4 is a one to one function
Crystal
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
can you show the steps from going from 3/(x-2)= y to x= 3/y +2 I'm confused as to how y ends up as the divisor
step 1: take reciprocal of both sides (x-2)/3 = 1/y step 2: multiply both sides by 3 x-2 = 3/y step 3: add 2 to both sides x = 3/y + 2 ps nice farcry 3 background!
Andrew
first you cross multiply and get y(x-2)=3 then apply distribution and the left side of the equation such as yx-2y=3 then you add 2y in both sides of the equation and get yx=3+2y and last divide both sides of the equation by y and you get x=3/y+2
Ioana
Multiply both sides by (x-2) to get 3=y(x-2) Then you can divide both sides by y (it's just a multiplied term now) to get 3/y = (x-2). Since the parentheses aren't doing anything for the right side, you can drop them, and add the 2 to both sides to get 3/y + 2 = x
Melin
thank you ladies and gentlemen I appreciate the help!
Robert
keep practicing and asking questions, practice makes perfect! and be aware that are often different paths to the same answer, so the more you familiarize yourself with these multiple different approaches, the less confused you'll be.
Andrew
please how do I learn integration
they are simply "anti-derivatives". so you should first learn how to take derivatives of any given function before going into taking integrals of any given function.
Andrew
best way to learn is always to look into a few basic examples of different kinds of functions, and then if you have any further questions, be sure to state specifically which step in the solution you are not understanding.
Andrew
example 1) say f'(x) = x, f(x) = ? well there is a rule called the 'power rule' which states that if f'(x) = x^n, then f(x) = x^(n+1)/(n+1) so in this case, f(x) = x^2/2
Andrew
great noticeable direction
Isaac
limit x tend to infinite xcos(π/2x)*sin(π/4x)
can you give me a problem for function. a trigonometric one