Week 5 — Review Questions (Part 2)

Exponential and trig derivatives, the product rule, and the quotient rule

Question 1 — Exponential Plus a Power

Find the derivative of

\begin{equation*} f(x) = 13e^x + x^6 \end{equation*}

Solution 1

The derivative of \(e^x\) is \(e^x\text{,}\) and the power rule handles \(x^6\text{:}\)

\begin{equation*} f'(x) = 13e^x + 6x^5 \end{equation*}

Question 2 — Exponential vs. Power

Find the derivative of

\begin{equation*} y(x) = d^x + x^d \end{equation*}

where \(d\) is a positive constant.

Solution 2

\(d^x\) is an exponential function (variable in the exponent), while \(x^d\) is a power function (constant exponent):

\begin{equation*} \dfrac{d}{dx}\left[d^x\right] = d^x \ln d \qquad \dfrac{d}{dx}\left[x^d\right] = d\,x^{d-1} \end{equation*}

\begin{equation*} y'(x) = d^x \ln d + d\,x^{d-1} \end{equation*}

Question 3 — Rate of Population Change

Since January 1, 1960, the population of Slim Chance has been described by

\begin{equation*} P = 36000(0.9)^t \end{equation*}

where \(t\) is years after the start of 1960. At what rate was the population changing on January 1, 1985?

Solution 3

Differentiate using \(\frac{d}{dt}\left[a^t\right] = a^t \ln a\text{:}\)

\begin{equation*} P'(t) = 36000(0.9)^t \ln(0.9) \end{equation*}

January 1, 1985 is \(t = 25\text{:}\)

\begin{equation*} P'(25) = 36000(0.9)^{25}\ln(0.9) \approx -272.3 \text{ people/yr} \end{equation*}

The population was decreasing by about 272 people per year.

Question 4 — Tangent and Normal Lines

Consider \(f(x) = 8 - e^x\text{.}\)

  1. Find the slope of the graph of \(f\) at the point where the graph crosses the \(x\)-axis.
  2. Find the equation of the tangent line at this point.
  3. Find the equation of the normal line (perpendicular to the tangent line) at this point.

Solution 4

\(f(x) = 0\) when \(e^x = 8\text{,}\) so \(x = \ln 8\text{.}\) Since \(f'(x) = -e^x\text{,}\)

\begin{equation*} \text{A.}\quad f'(\ln 8) = -e^{\ln 8} = -8 \end{equation*}

\begin{equation*} \text{B.}\quad y = -8(x - \ln 8) = -8x + 8\ln 8 \end{equation*}

The normal slope is the negative reciprocal, \(\frac{1}{8}\text{:}\)

\begin{equation*} \text{C.}\quad y = \dfrac{1}{8}(x - \ln 8) = \dfrac{1}{8}x - \dfrac{\ln 8}{8} \end{equation*}

Question 5 — Product Rule with an Exponential

Find the derivative of

\begin{equation*} f(x) = \left(x^6 - \sqrt[5]{x}\right)4^x \end{equation*}

Solution 5

Rewrite \(\sqrt[5]{x} = x^{1/5}\) and use the product rule \((uv)' = u'v + uv'\text{:}\)

\begin{equation*} u = x^6 - x^{1/5}, \quad u' = 6x^5 - \dfrac{1}{5}x^{-4/5} \qquad v = 4^x, \quad v' = 4^x \ln 4 \end{equation*}

\begin{equation*} f'(x) = \left(6x^5 - \dfrac{1}{5}x^{-4/5}\right)4^x + \left(x^6 - x^{1/5}\right)4^x \ln 4 \end{equation*}

Question 6 — Quotient Rule

Find the derivative of

\begin{equation*} z = \dfrac{9t + 1}{4t + 4} \end{equation*}

Solution 6

Quotient rule, \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\text{:}\)

\begin{equation*} \dfrac{dz}{dt} = \dfrac{9(4t + 4) - (9t + 1)(4)}{(4t + 4)^2} = \dfrac{36t + 36 - 36t - 4}{(4t + 4)^2} \end{equation*}

\begin{equation*} \dfrac{dz}{dt} = \dfrac{32}{(4t + 4)^2} = \dfrac{2}{(t + 1)^2} \end{equation*}

Question 7 — Product and Quotient Rules from Values

Let

\begin{equation*} F(2) = 5, \quad F'(2) = 3, \quad H(2) = 5, \quad H'(2) = 3 \end{equation*}

  1. If \(G(z) = F(z) \cdot H(z)\text{,}\) find \(G'(2)\text{.}\)
  2. If \(G(w) = F(w) / H(w)\text{,}\) find \(G'(2)\text{.}\)

Solution 7

\begin{equation*} \text{A.}\quad G'(2) = F'(2)H(2) + F(2)H'(2) = 3(5) + 5(3) = 30 \end{equation*}

\begin{equation*} \text{B.}\quad G'(2) = \dfrac{F'(2)H(2) - F(2)H'(2)}{H(2)^2} = \dfrac{15 - 15}{25} = 0 \end{equation*}

Question 8 — Product Rule from a Graph

The graph of \(f(x)\) is black and has a sharp corner at \(x = 2\text{.}\) The graph of \(g(x)\) is blue.

Let \(h(x) = f(x) \cdot g(x)\text{.}\) Find the following, or state that the derivative does not exist.

  1. \(\displaystyle h'(1)\)
  2. \(\displaystyle h'(2)\)
  3. \(\displaystyle h'(3)\)

Graph on a grid from -1 to 4 in both directions. The black graph of f goes in straight segments from (0,0) up to a sharp corner at (2,2), then down to (4,0). The blue graph of g is a straight line from (0,3) to (4,0).

Solution 8

From the graph: \(g(x) = 3 - \frac{3}{4}x\text{,}\) so \(g' = -\frac{3}{4}\text{.}\) \(f\) has slope \(1\) on \((0,2)\) and \(-1\) on \((2,4)\text{.}\) Use \(h' = f'g + fg'\text{.}\)

\begin{equation*} \text{A.}\quad h'(1) = (1)\left(\tfrac{9}{4}\right) + (1)\left(-\tfrac{3}{4}\right) = \tfrac{3}{2} \end{equation*}

\begin{equation*} \text{B.}\quad h'(2) \text{ does not exist } (f \text{ has a corner at } x = 2) \end{equation*}

\begin{equation*} \text{C.}\quad h'(3) = (-1)\left(\tfrac{3}{4}\right) + (1)\left(-\tfrac{3}{4}\right) = -\tfrac{3}{2} \end{equation*}

Question 9 — Product Rule with Trig

Find the derivative of

\begin{equation*} h(t) = t\sin t + \cos t \end{equation*}

Solution 9

Product rule on \(t\sin t\text{,}\) and \(\frac{d}{dt}[\cos t] = -\sin t\text{:}\)

\begin{equation*} h'(t) = (1)\sin t + t\cos t - \sin t = t\cos t \end{equation*}

Question 10 — Tangent Line to a Trig Curve

Find the equation of the tangent line to the curve

\begin{equation*} y = 3\sec(x) - 6\cos(x) \end{equation*}

at the point \(\left(\frac{\pi}{3}, 3\right)\text{.}\) Write your answer in the form \(y = mx + b\text{.}\)

Solution 10

\begin{equation*} y' = 3\sec x \tan x + 6\sin x \end{equation*}

At \(x = \frac{\pi}{3}\text{:}\) \(\sec = 2\text{,}\) \(\tan = \sqrt{3}\text{,}\) \(\sin = \frac{\sqrt{3}}{2}\text{.}\)

\begin{equation*} m = 3(2)(\sqrt{3}) + 6\left(\tfrac{\sqrt{3}}{2}\right) = 9\sqrt{3} \end{equation*}

\begin{equation*} y - 3 = 9\sqrt{3}\left(x - \tfrac{\pi}{3}\right) \quad\Rightarrow\quad y = 9\sqrt{3}\,x - 3\sqrt{3}\,\pi + 3 \end{equation*}