Week 5 — Chain Rule

Composite functions, nested chain rule, and chain rule with the product rule

Question 1 — Power of a Polynomial

Differentiate the following function.

\begin{equation*} y = \left(3x^2 - 5\right)^4 \end{equation*}

Solution 1

Outer function \(u^4\text{,}\) inner function \(u = 3x^2 - 5\text{:}\)

\begin{equation*} \dfrac{dy}{dx} = 4\left(3x^2 - 5\right)^3 \cdot 6x = 24x\left(3x^2 - 5\right)^3 \end{equation*}

Question 2 — Exponential of a Polynomial

Differentiate the following function.

\begin{equation*} y = e^{x^3 - 2x} \end{equation*}

Solution 2

The derivative of \(e^u\) is \(e^u \cdot u'\text{,}\) with \(u = x^3 - 2x\text{:}\)

\begin{equation*} \dfrac{dy}{dx} = \left(3x^2 - 2\right)e^{x^3 - 2x} \end{equation*}

Question 3 — An Exponential Inside an Exponential

Differentiate the following function.

\begin{equation*} y = e^{e^{x^4}} \end{equation*}

Solution 3

Apply the chain rule twice, working from the outside in:

\begin{equation*} \dfrac{dy}{dx} = e^{e^{x^4}} \cdot \dfrac{d}{dx}\left[e^{x^4}\right] = e^{e^{x^4}} \cdot e^{x^4} \cdot 4x^3 \end{equation*}

\begin{equation*} \dfrac{dy}{dx} = 4x^3 e^{x^4} e^{e^{x^4}} \end{equation*}

Question 4 — \(\sin(x^2)\) vs. \(\sin^2 x\)

Differentiate the following function.

\begin{equation*} y = \sin\left(x^2\right) + \sin^2 x \end{equation*}

Solution 4

In \(\sin(x^2)\) the inside is \(x^2\text{.}\) In \(\sin^2 x = (\sin x)^2\) the inside is \(\sin x\text{:}\)

\begin{equation*} \dfrac{d}{dx}\left[\sin\left(x^2\right)\right] = \cos\left(x^2\right)\cdot 2x \qquad \dfrac{d}{dx}\left[(\sin x)^2\right] = 2\sin x \cos x \end{equation*}

\begin{equation*} \dfrac{dy}{dx} = 2x\cos\left(x^2\right) + 2\sin x\cos x \end{equation*}

Question 5 — Trig Inside Trig

Find the derivative of

\begin{equation*} g(q) = \cos(\tan q) \end{equation*}

Solution 5

Outer function \(\cos u\text{,}\) inner function \(u = \tan q\text{:}\)

\begin{equation*} g'(q) = -\sin(\tan q)\cdot \sec^2 q \end{equation*}

Question 6 — Three Layers

Differentiate the following function.

\begin{equation*} y = \cos^3(4x) \end{equation*}

Solution 6

Rewrite as \(y = \left(\cos(4x)\right)^3\text{.}\) Layers: cube, cosine, \(4x\text{:}\)

\begin{equation*} \dfrac{dy}{dx} = 3\cos^2(4x)\cdot\left(-\sin(4x)\right)\cdot 4 \end{equation*}

\begin{equation*} \dfrac{dy}{dx} = -12\cos^2(4x)\sin(4x) \end{equation*}

Question 7 — Product of Trig Powers

Find the derivative of

\begin{equation*} k(a) = \sin^3 a \cos^6 a \end{equation*}

Solution 7

Product rule, with the chain rule on each factor:

\begin{equation*} k'(a) = 3\sin^2 a\cos a\cdot\cos^6 a + \sin^3 a\cdot 6\cos^5 a\,(-\sin a) \end{equation*}

\begin{equation*} k'(a) = 3\sin^2 a\cos^7 a - 6\sin^4 a\cos^5 a \end{equation*}

\begin{equation*} k'(a) = 3\sin^2 a\cos^5 a\left(\cos^2 a - 2\sin^2 a\right) \end{equation*}

Question 8 — Chain Rule from Given Values

Let \(F(x) = f\left(x^7\right)\) and \(G(x) = \left(f(x)\right)^7\text{.}\) You also know that

\begin{equation*} a^6 = 12, \quad f(a) = 3, \quad f'(a) = 11, \quad f'\left(a^7\right) = 6 \end{equation*}

Find \(F'(a)\) and \(G'(a)\text{.}\)

Solution 8

For \(F\text{,}\) the inside is \(x^7\text{:}\)

\begin{equation*} F'(x) = f'\left(x^7\right)\cdot 7x^6 \quad\Rightarrow\quad F'(a) = 6\cdot 7\cdot 12 = 504 \end{equation*}

For \(G\text{,}\) the inside is \(f(x)\text{:}\)

\begin{equation*} G'(x) = 7\left(f(x)\right)^6 f'(x) \quad\Rightarrow\quad G'(a) = 7\cdot 3^6\cdot 11 = 56133 \end{equation*}

Question 9 — Product Rule with a Root

Differentiate the following function.

\begin{equation*} y = x^2\sqrt{1 - 3x} \end{equation*}

Solution 9

Product rule, with the chain rule on \((1 - 3x)^{1/2}\text{:}\)

\begin{equation*} \dfrac{dy}{dx} = 2x\sqrt{1 - 3x} + x^2\cdot\dfrac{-3}{2\sqrt{1 - 3x}} \end{equation*}

Combine over a common denominator:

\begin{equation*} \dfrac{dy}{dx} = \dfrac{4x(1 - 3x) - 3x^2}{2\sqrt{1 - 3x}} = \dfrac{x(4 - 15x)}{2\sqrt{1 - 3x}} \end{equation*}

Question 10 — Damped Spring

The position of a point on a damped spring is

\begin{equation*} s(t) = 8e^{-1.1t}\sin(2\pi t) \end{equation*}

where \(s\) is in centimeters and \(t\) is in seconds. Find the velocity of the point after \(t\) seconds.

Solution 10

Velocity is \(s'(t)\text{.}\) Product rule, with the chain rule on each factor:

\begin{equation*} v(t) = 8\left[-1.1e^{-1.1t}\sin(2\pi t) + e^{-1.1t}\cdot 2\pi\cos(2\pi t)\right] \end{equation*}

\begin{equation*} v(t) = e^{-1.1t}\left(16\pi\cos(2\pi t) - 8.8\sin(2\pi t)\right) \end{equation*}