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You CAN Ace Calculus

17calculus > derivatives > chain rule

Topics You Need To Understand For This Page

basic derivative rules

power rule

product rule

quotient rule

Some of the practice problems require you to know one or more of the following rules. If you haven't learned all these rules yet, no worries. You can filter out the problems that require techniques you do not know.

trig derivatives

exponential derivatives

derivatives of logarithms

Calculus Main Topics

Tools

Related Topics and Links

Chain Rule FAQs

ATTENTION INSTRUCTORS: The new 2018 version of 17calculus will include changes to the practice problem numbering system. If you would like advance information to help you prepare for spring semester, send us an email at 2018info at 17calculus.com.

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Derivative - Chain Rule

on this page: ► substitution method     ► direct method

The Chain Rule is probably the most important derivative rule that you will learn since you will need to use it a lot and it shows up in various forms in other derivatives and integration. It can also be a little confusing at first but if you stick with it, you will be able to understand it well.

There are two main ways to learn the chain rule, substitution and direct. I recommend writing out the substitution form while you are first learning. After you get the hang of it, you can use the direct (shorter) version.
NOTE: Do not be lazy here and try to use the direct version first. Write everything out so that you can see it. If you don't, you will struggle with this concept for the rest of your calculus life and this rule is used everywhere, and I mean EVERYWHERE from now on. You can't get away from it and you can't get around it. So, let's get started.

What The Chain Rule Says

The chain rule says, if you have a function in the form y=f(u) where u is a function of x, then \(\displaystyle{ \frac{dy}{dx} = \frac{df}{du} \cdot \frac{du}{dx} }\).
The notation tells you that \(u(x)\) is a composite function of \(f\).

Substitution Method

This is the method you should learn first since it breaks the problem into smaller, bite-size pieces that you already know how to take derivatives of. The best way to learn this concept is with examples.

1. \( y=(x+1)^2 \) can be separated into \(u = x+1 \) and \( y=u^2 \). We started on the outside and noticed that there was something with a power of 2. So we took that something and set it equal to u.

2. \( y=\sin(x^2) \) can be separated into \( u=x^2 \) and \( y=\sin(u) \). We noticed that, on the outside, we have the sine function. So we set whatever was on the inside to u.

3. \( y=\sqrt{\sin(x)} \) has an outside function as a square root, while the inside function is sine. So \( u=\sin(x) \) and \( y = \sqrt{u} \).

4. \( y=\sqrt{\sin(x^2)} \) is a combination of the last two examples and is a little more complicated (but not much). We do the same thing as before and notice that the outside function a square root. So we set u to the inside function, i.e. \( u=\sin(x^2) \) and \( y=\sqrt{u} \). When we try to take the derivative \(du/dx\), we need to do substitution again. So we let \(v=x^2\) and \(u=\sin(v)\).

In all of these examples, we just showed you how to set the problem with substitution. You would then complete the problem by taking the derivative of each piece. Try this example before going on to the direct method.

Example - - Calculate the derivative of \( y=\cos(x^2+3) \) using the substitution method.

Direct Method

Okay, once you get the basic idea of the chain rule down, the next step is be able to write out the derivative without using substitution. Note: Some instructors jump to this technique right away without emphasizing the substitution concept. If your instructor does that, check your textbook and learn the chain rule with substitution first. You will be glad you did later on down the road.

Now, there are two ways to think about working problems without substitution. The technique that I think works best is to start from the outside and work your way in. Let's see how this is done using an example of an outside to inside technique.

Example - - Calculate the derivative of \( g(t)=\sin(t^3) \) using the direct method.

A second way to work this is to start on the inside and work your way out. Some of the videos further down on this page show this technique but we do not recommend it. The reason is that the more difficult problems (and ones you will probably see on your exam) are nested, i.e. you need to use the chain rule repeatedly. It is much easier to break the problem down and see what you need to do if you start on the outside and work your way in. Starting in the inside is sometimes difficult to find where to start. Let's do an example of this.

Example - - Evaluate \( \displaystyle{ \frac{d}{dx}[\sin(\tan(2x))]}\).

Before working practice problems, let's watch a video. This video has a great explanation of the chain rule at the first and then some examples that should help you understand how to use this rule.

PatrickJMT - Chain Rule for Finding Derivatives

Okay, now you are ready for some practice problems.

Search 17Calculus

practice filters

use basic derivatives only (9)

use trig rules (9)

use exponential and/or logarithmic rules (7)

Practice Problems

Instructions - - Unless otherwise instructed, calculate the derivative of these functions and give your final answers in completely factored form.

Level A - Basic

Practice A01

\(\displaystyle{ 3e^{ x^2+7 } }\)

answer

solution

Practice A02

\(\displaystyle{ (x^3+1)^4 }\)

answer

solution

Practice A03

\(\displaystyle{ \ln(3x^2+9x-5) }\)

answer

solution

Practice A04

Calculate the first three derivatives of \(f(x)=\ln(2+3x)\).

solution

Practice A05

\(\displaystyle{ f(x)=\sqrt{2x+1} }\)

answer

solution

Practice A06

\(\displaystyle{ f(x)=(x+1)^3 }\)

solution

Practice A07

\(\displaystyle{ 7 \sin(x^2+1) }\)

answer

solution

Practice A08

\(\displaystyle{ f(x) = (x^4+3x^2-2)^5 }\)

solution

Practice A09

\( \tan(x^5 + 2x^3 - 12x) \)

answer

solution

Practice A10

\(y=4^{x^3-\sin x}\)

answer

solution

Practice A11

\(\displaystyle{ g(x)= (1+4x)^5(3+x-x^2)^8 }\)

solution

Practice A12

\(\displaystyle{ f(x) = (2x-3)^4 (x^2+x+1)^5 }\)

solution

Practice A13

\(\displaystyle{ y=\sin^3(x) \tan(4x) }\)

solution

Practice A14

\(\displaystyle{ y = (2x^5-3) \sin(7x) }\)

answer

solution

Practice A15

\(y=x^2\cos(1/x^3)\)

answer

solution

Practice A16

\((x^2-1)e^{-x}\)

answer

solution


Level B - Intermediate

Practice B01

\(\displaystyle{ y = (1+\cos^2(7x))^3 }\)

solution

Practice B02

\(\displaystyle{ y = \cos \left( \frac{1-e^{2x}}{1+e^{2x}} \right) }\)

solution

Practice B03

\(\displaystyle{ y=\left( \frac{x^2+1}{x^2-1} \right)^3 }\)

solution

Practice B04

\(\displaystyle{ g(t) = \frac{(t+4)^{1/2}}{(t-4)^{1/2}} }\)

answer

solution

Practice B05

Use the quotient rule to find \(f'(x)\) of \(\displaystyle{ f(x)=\frac{6}{\ln(8x^2)} }\).

answer

solution

Practice B06

\(\displaystyle{y=\ln\sqrt{\frac{x^2+1}{x+3}}}\)

answer

solution


Level C - Advanced

Practice C01

\(\displaystyle{ y=\left[ \tan \left( \sin \left( \sqrt{x^2+8x}\right) \right) \right]^5 }\)

solution

Practice C02

\(\displaystyle{ y=x \sin(1/x) + \sqrt[4]{(1-3x)^2 + x^5} }\)

solution

Practice C03

\(\displaystyle{ f(x)=\left[ \frac{x+4}{\sqrt{x^2+1}} \right]^3 }\)

solution

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