MAT147 — Week 5: Chain Rule (Part 2)

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Week 5 — Chain Rule (Part 2)

More practice with the chain rule

Question 1 — A Power in the Denominator

Differentiate the following function.

\begin{equation*} y = \dfrac{1}{\left(x^2 + 4\right)^3} \end{equation*}

Solution 1

Rewrite with a negative exponent:

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

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

Question 2 — Tangent of an Exponential

Differentiate the following function.

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

Solution 2

Outer function $\tan u\text{,}$ inner function $u = e^{2x}\text{,}$ which itself needs the chain rule:

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

Question 3 — Rate of Temperature Change

The temperature (in degrees) $t$ hours after midnight is modeled by

\begin{equation*} T(t) = 60 + 15\sin\left(\dfrac{\pi t}{12}\right) \end{equation*}

Find the rate of change of the temperature at $t = 4\text{.}$

Solution 3

\begin{equation*} T'(t) = 15\cos\left(\dfrac{\pi t}{12}\right)\cdot\dfrac{\pi}{12} = \dfrac{5\pi}{4}\cos\left(\dfrac{\pi t}{12}\right) \end{equation*}

At $t = 4\text{,}$ $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\text{:}$

\begin{equation*} T'(4) = \dfrac{5\pi}{8} \approx 1.96 \text{ degrees per hour} \end{equation*}

Question 4 — Product of Two Powers

Differentiate the following function.

\begin{equation*} y = \left(x^2 + 1\right)^3\left(2x - 1\right)^4 \end{equation*}

Solution 4

Product rule, with the chain rule on each factor:

\begin{equation*} \dfrac{dy}{dx} = 3\left(x^2 + 1\right)^2(2x)\left(2x - 1\right)^4 + \left(x^2 + 1\right)^3\cdot 4\left(2x - 1\right)^3(2) \end{equation*}

Factor out $2\left(x^2 + 1\right)^2\left(2x - 1\right)^3\text{:}$

\begin{equation*} \dfrac{dy}{dx} = 2\left(x^2 + 1\right)^2\left(2x - 1\right)^3\left(10x^2 - 3x + 4\right) \end{equation*}

Question 5 — Chain Rule from Graphs

Use the graphs of $f$ and $g$ to find:

$h'(1)$ if $h(x) = f(g(x))$
$k'(2)$ if $k(x) = g(f(x))$

Solution 5

From the graphs: $g(1) = 3\text{,}$ $g'(1) = 2\text{,}$ $f'(3) = -1\text{,}$ and $f(2) = 3\text{,}$ $f'(2) = -1\text{,}$ $g'(3) = -1\text{.}$

\begin{equation*} h'(1) = f'(g(1))\cdot g'(1) = f'(3)\cdot 2 = (-1)(2) = -2 \end{equation*}

\begin{equation*} k'(2) = g'(f(2))\cdot f'(2) = g'(3)\cdot(-1) = (-1)(-1) = 1 \end{equation*}

Question 6 — Cosine of a Root

Differentiate the following function.

\begin{equation*} y = \cos\left(\sqrt{x}\right) \end{equation*}

Solution 6

Outer function $\cos u\text{,}$ inner function $u = x^{1/2}\text{:}$

\begin{equation*} \dfrac{dy}{dx} = -\sin\left(\sqrt{x}\right)\cdot\dfrac{1}{2\sqrt{x}} = -\dfrac{\sin\left(\sqrt{x}\right)}{2\sqrt{x}} \end{equation*}

Question 7 — Working Backwards to f′

\begin{equation*} \dfrac{d}{dx}\left(f\left(2x^3\right)\right) = 12x^8 \end{equation*}

calculate $f'(x)\text{.}$

Solution 7

By the chain rule:

\begin{equation*} f'\left(2x^3\right)\cdot 6x^2 = 12x^8 \quad\Rightarrow\quad f'\left(2x^3\right) = 2x^6 \end{equation*}

Let $u = 2x^3\text{,}$ so $x^3 = \frac{u}{2}$ and $x^6 = \frac{u^2}{4}\text{:}$

\begin{equation*} f'(u) = 2\cdot\dfrac{u^2}{4} = \dfrac{u^2}{2} \quad\Rightarrow\quad f'(x) = \dfrac{x^2}{2} \end{equation*}

Question 8 — Exponential with a Constant

Differentiate the following function, where $k$ is a constant.

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

Solution 8

Inner function $u = -kx^2\text{,}$ so $u' = -2kx\text{:}$

\begin{equation*} \dfrac{dy}{dx} = 3e^{-kx^2}\cdot(-2kx) = -6kx\,e^{-kx^2} \end{equation*}

Question 9 — Tangent Line

Find the equation of the tangent line to

\begin{equation*} y = \sqrt{x^2 + 9} \end{equation*}

at $x = 4\text{.}$

Solution 9

Point: $y(4) = \sqrt{25} = 5\text{.}$

\begin{equation*} y' = \dfrac{1}{2\sqrt{x^2 + 9}}\cdot 2x = \dfrac{x}{\sqrt{x^2 + 9}} \quad\Rightarrow\quad m = \dfrac{4}{5} \end{equation*}

\begin{equation*} y - 5 = \dfrac{4}{5}(x - 4) \quad\Rightarrow\quad y = \dfrac{4}{5}x + \dfrac{9}{5} \end{equation*}

Question 10 — Product with an Exponential

Differentiate the following function.

\begin{equation*} y = \sin x\cdot e^{\cos x} \end{equation*}

Solution 10

Product rule, with the chain rule on $e^{\cos x}\text{:}$

\begin{equation*} \dfrac{dy}{dx} = \cos x\cdot e^{\cos x} + \sin x\cdot e^{\cos x}(-\sin x) \end{equation*}

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

Question 11 — Look for an Identity

Differentiate the following function.

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

Solution 11

Use $1 - \sin^2 x = \cos^2 x\text{:}$

\begin{equation*} y = \sqrt{\cos^2 x} = \left|\cos x\right| \end{equation*}

Where $\cos x \gt 0\text{:}$ $y = \cos x\text{,}$ so $y' = -\sin x\text{.}$

Where $\cos x \lt 0\text{:}$ $y = -\cos x\text{,}$ so $y' = \sin x\text{.}$