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In exercises 1 - 6, given \(y=f(u)\) and \(u=g(x)\), find \(\dfrac{dy}{dx}\) by using Leibniz’s notation for the chain rule: \(\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}.\)
1) \(y=3u−6,\quad u=2x^2\)
2) \(y=6u^3,\quad u=7x−4\)
- Answer
- \(\dfrac{dy}{dx} = 18u^2⋅7=18(7x−4)^2⋅7= 126(7x−4)^2\)
3) \(y=\sin u,\quad u=5x−1\)
4) \(y=\cos u,\quad u=-\frac{x}{8}\)
- Answer
- \(\dfrac{dy}{dx} = −\sin u⋅\left(-\frac{1}{8}\right)=\frac{1}{8}\sin(-\frac{x}{8})\)
5) \(y=\tan u,\quad u=9x+2\)
6) \(y=\sqrt{4u+3},\quad u=x^2−6x\)
- Answer
- \(\dfrac{dy}{dx} = \dfrac{8x−24}{2\sqrt{4u+3}}=\dfrac{4x−12}{\sqrt{4x^2−24x+3}}\)
For each of the following exercises,
a. decompose each function in the form \(y=f(u)\) and \(u=g(x)\), and
b. find \(\dfrac{dy}{dx}\) as a function of \(x\).
7) \(y=(3x−2)^6\)
8) \(y=(3x^2+1)^3\)
- Answer
- a. \(f(u)=u^3,\quad u=3x^2+1\);
b. \(\dfrac{dy}{dx} = 18x(3x^2+1)^2\)
9) \(y=\sin^5(x)\)
10) \(y=\left(\dfrac{x}{7}+\dfrac{7}{x}\right)^7\)
- Answer
- a. \(f(u)=u^7,\quad u=\dfrac{x}{7}+\dfrac{7}{x}\);
b. \(\dfrac{dy}{dx} = 7\left(\dfrac{x}{7}+\dfrac{7}{x}\right)^6⋅\left(\dfrac{1}{7}−\dfrac{7}{x^2}\right)\)
11) \(y=\tan(\sec x)\)
12) \(y=\csc(πx+1)\)
- Answer
- a. \(f(u)=\csc u,\quad u=πx+1\);
b. \(\dfrac{dy}{dx} = −π\csc(πx+1)⋅\cot(πx+1)\)
13) \(y=\cot^2x\)
14) \(y=−6\sin^{−3}x\)
- Answer
- a. \(f(u)=−6u^{−3},\quad u=\sin x\);
b. \(\dfrac{dy}{dx} = 18\sin^{−4}x⋅\cos x\)
In exercises 15 - 24, find \(\dfrac{dy}{dx}\) for each function.
15) \(y=(3x^2+3x−1)^4\)
16) \(y=(5−2x)^{−2}\)
- Answer
- \(\dfrac{dy}{dx}=\dfrac{4}{(5−2x)^3}\)
17) \(y=\cos^3(πx)\)
18) \(y=(2x^3−x^2+6x+1)^3\)
- Answer
- \(\dfrac{dy}{dx}=6(2x^3−x^2+6x+1)^2⋅(3x^2−x+3)\)
19) \(y=\dfrac{1}{\sin^2(x)}\)
20) \(y=\big(\tan x+\sin x\big)^{−3}\)
- Answer
- \(\dfrac{dy}{dx}=−3\big(\tan x+\sin x\big)^{−4}⋅(\sec^2x+\cos x)\)
21) \(y=x^2\cos^4x\)
22) \(y=\sin(\cos 7x)\)
- Answer
- \(\dfrac{dy}{dx}=−7\cos(\cos 7x)⋅\sin 7x\)
23) \(y=\sqrt{6+\sec πx^2}\)
24) \(y=\cot^3(4x+1)\)
- Answer
- \(\dfrac{dy}{dx}=−12\cot^2(4x+1)⋅\csc^2(4x+1)\)
25) Let \(y=\big[f(x)\big]^3\) and suppose that \(f′(1)=4\) and \(\frac{dy}{dx}=10\) for \(x=1\). Find \(f(1)\).
26) Let \(y=\big(f(x)+5x^2\big)^4\) and suppose that \(f(−1)=−4\) and \(\frac{dy}{dx}=3\) when \(x=−1\). Find \(f′(−1)\)
- Answer
- \(f′(−1)=10\frac{3}{4}\)
27) Let \(y=(f(u)+3x)^2\) and \(u=x^3−2x\). If \(f(4)=6\) and \(\frac{dy}{dx}=18\) when \(x=2\), find \(f′(4)\).
28) [T] Find the equation of the tangent line to \(y=−\sin(\frac{x}{2})\) at the origin. Use a calculator to graph the function and the tangent line together.
- Answer
- \(y=-\frac{1}{2}x\)
29) [T] Find the equation of the tangent line to \(y=\left(3x+\frac{1}{x}\right)^2\) at the point \((1,16)\). Use a graphing calculator or Desmos to graph the function and the tangent line together.
30) Find the \(x\) -coordinates at which the tangent line to \(y=\left(x−\frac{6}{x}\right)^8\) is horizontal.
- Answer
- \(x=±\sqrt{6}\)
31) [T] Find an equation of the line that is normal to \(g(θ)=\sin^2(πθ)\) at the point \(\left(\frac{1}{4},\frac{1}{2}\right)\). Use a calculator to graph the function and the normal line together.
For exercises 32 - 39, use the information in the following table to find \(h′(a)\) at the given value for \(a\).
| \(x\) | \(f(x)\) | \(f'(x)\) | \(g(x)\) | \(g'(x)\) |
| 0 | 2 | 5 | 0 | 2 |
| 1 | 1 | −2 | 3 | 0 |
| 2 | 4 | 4 | 1 | −1 |
| 3 | 3 | −3 | 2 | 3 |
32) \(h(x)=f\big(g(x)\big);\quad a=0\)
- Answer
- \(h'(0) = 10\)
33) \(h(x)=g\big(f(x)\big);\quad a=0\)
34) \(h(x)=\big(x^4+g(x)\big)^{−2};\quad a=1\)
- Answer
- \(h'(1) = −\frac{1}{8}\)
35) \(h(x)=\left(\dfrac{f(x)}{g(x)}\right)^2;\quad a=3\)
36) \(h(x)=f\big(x+f(x)\big);\quad a=1\)
- Answer
- \(h'(1) = −4\)
37) \(h(x)=\big(1+g(x)\big)^3;\quad a=2\)
38) \(h(x)=g\big(2+f(x^2)\big);\quad a=1\)
- Answer
- \(h'(1) = −12\)
39) \(h(x)=f\big(g(\sin x)\big);\quad a=0\)
This page titled 3.4E: Exercises for Section 3.4 was last modified on Wed, 12 Aug 2026 20:53:31 GMT and is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Leonard Lamp, Kenn Huber, and Mohamad Youssef.
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