Week 6 — Review Questions

Logarithms, inverse trig, and implicit differentiation

Question 1 — Chain Rule with ln

Differentiate the following function.

\begin{equation*} f(t) = \ln\left(t^7 + 7\right) \end{equation*}

Solution 1

Use \(\frac{d}{dt}[\ln u] = \frac{u'}{u}\text{:}\)

\begin{equation*} f'(t) = \dfrac{7t^6}{t^7 + 7} \end{equation*}

Question 2 — Sine of a Logarithm

Differentiate the following function.

\begin{equation*} g(t) = \sin(\ln t) \end{equation*}

Solution 2

Chain rule, with \(\frac{d}{dt}[\ln t] = \frac{1}{t}\text{:}\)

\begin{equation*} g'(t) = \cos(\ln t)\cdot\dfrac{1}{t} = \dfrac{\cos(\ln t)}{t} \end{equation*}

Question 3 — Product Rule with ln

Differentiate the following function.

\begin{equation*} y = 2x\left(\ln x + \ln 7\right) - 5x + \pi \end{equation*}

Solution 3

Product rule on the first term; \(\ln 7\) and \(\pi\) are constants:

\begin{equation*} \dfrac{dy}{dx} = 2\left(\ln x + \ln 7\right) + 2x\cdot\dfrac{1}{x} - 5 \end{equation*}

\begin{equation*} \dfrac{dy}{dx} = 2\ln x + 2\ln 7 - 3 \end{equation*}

Question 4 — Inverse Sine

Differentiate the following function.

\begin{equation*} f(x) = 3\sin^{-1}\left(x^4\right) \end{equation*}

Solution 4

Use \(\frac{d}{dx}[\sin^{-1} u] = \frac{u'}{\sqrt{1-u^2}}\) with \(u = x^4\text{:}\)

\begin{equation*} f'(x) = 3\cdot\dfrac{4x^3}{\sqrt{1 - x^8}} = \dfrac{12x^3}{\sqrt{1 - x^8}} \end{equation*}

Question 5 — Product Rule with Inverse Tangent

Differentiate the following function.

\begin{equation*} f(x) = 9x^4\arctan\left(5x^4\right) \end{equation*}

Solution 5

Product rule, with \(\frac{d}{dx}[\arctan u] = \frac{u'}{1+u^2}\) and \(u = 5x^4\text{:}\)

\begin{equation*} f'(x) = 36x^3\arctan\left(5x^4\right) + 9x^4\cdot\dfrac{20x^3}{1 + 25x^8} \end{equation*}

\begin{equation*} f'(x) = 36x^3\arctan\left(5x^4\right) + \dfrac{180x^7}{1 + 25x^8} \end{equation*}

Question 6 — Cosine Times Inverse Sine

Differentiate the following function.

\begin{equation*} f(x) = 4\cos(x)\sin^{-1}(x) \end{equation*}

Solution 6

Product rule:

\begin{equation*} f'(x) = -4\sin(x)\sin^{-1}(x) + \dfrac{4\cos(x)}{\sqrt{1 - x^2}} \end{equation*}

Question 7 — Simplify First

Differentiate the following function.

\begin{equation*} y = \ln\left(x^3\right) - \ln\left(\dfrac{1}{x^2}\right) + 2\ln(x) - \ln\left(3x^7\right) \end{equation*}

Solution 7

Use log rules to simplify before differentiating:

\begin{equation*} y = 3\ln x + 2\ln x + 2\ln x - \ln 3 - 7\ln x = -\ln 3 = \ln\left(\tfrac{1}{3}\right) \end{equation*}

\(y\) is a constant, so

\begin{equation*} \dfrac{dy}{dx} = 0 \end{equation*}

Question 8 — Implicit Differentiation

Find \(\dfrac{dy}{dx}\) in terms of \(x\) and \(y\text{.}\)

\begin{equation*} x^4y - x - 9y - 8 = 0 \end{equation*}

Solution 8

Differentiate both sides with respect to \(x\text{:}\)

\begin{equation*} 4x^3y + x^4\dfrac{dy}{dx} - 1 - 9\dfrac{dy}{dx} = 0 \end{equation*}

\begin{equation*} \dfrac{dy}{dx}\left(x^4 - 9\right) = 1 - 4x^3y \quad\Rightarrow\quad \dfrac{dy}{dx} = \dfrac{1 - 4x^3y}{x^4 - 9} \end{equation*}

Question 9 — Slope on an Implicit Curve

Find the slope of the tangent to the curve

\begin{equation*} x^3 + 3xy + y^2 = 41 \end{equation*}

at the point \((1, 5)\text{.}\)

Solution 9

\begin{equation*} 3x^2 + 3y + 3x\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 0 \quad\Rightarrow\quad \dfrac{dy}{dx} = -\dfrac{3x^2 + 3y}{3x + 2y} \end{equation*}

At \((1, 5)\text{:}\)

\begin{equation*} \dfrac{dy}{dx} = -\dfrac{3 + 15}{3 + 10} = -\dfrac{18}{13} \end{equation*}

Question 10 — Tangent Line to an Ellipse

Find the equation of the tangent line to the ellipse

\begin{equation*} \dfrac{x^2}{25} + \dfrac{y^2}{16} = 1 \end{equation*}
at the point \(\left(3, \tfrac{16}{5}\right)\text{.}\)

An ellipse centered at the origin crossing the x-axis at plus and minus 5 and the y-axis at plus and minus 4, with the point (3, 16/5) marked in the first quadrant.

Solution 10

\begin{equation*} \dfrac{2x}{25} + \dfrac{2y}{16}\dfrac{dy}{dx} = 0 \quad\Rightarrow\quad \dfrac{dy}{dx} = -\dfrac{16x}{25y} \end{equation*}

At \(\left(3, \tfrac{16}{5}\right)\text{:}\) \(m = -\dfrac{48}{80} = -\dfrac{3}{5}\)

\begin{equation*} y - \dfrac{16}{5} = -\dfrac{3}{5}(x - 3) \quad\Rightarrow\quad y = -\dfrac{3}{5}x + 5 \end{equation*}