Week 5 — Review Questions

The power rule, tangent lines, and linear approximation

Question 1 — Differentiating a Polynomial

Differentiate the following function.

\begin{equation*} y = 2x^3 - 3x^2 + 6x \end{equation*}

Solution 1

Apply the power rule to each term:

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

Question 2 — A Negative Power

Differentiate the following function.

\begin{equation*} y = \dfrac{1}{x^{34}} \end{equation*}

Solution 2

Rewrite with a negative exponent:

\begin{equation*} y = x^{-34} \end{equation*}

\begin{equation*} \dfrac{dy}{dx} = -34x^{-35} = -\dfrac{34}{x^{35}} \end{equation*}

Question 3 — Radicals as Powers

Differentiate the following function.

\begin{equation*} u = \sqrt[3]{t^2} + 3\sqrt{t^3} \end{equation*}

Solution 3

Rewrite each radical as a fractional exponent:

\begin{equation*} u = t^{2/3} + 3t^{3/2} \end{equation*}

\begin{equation*} u' = \dfrac{2}{3}t^{-1/3} + 3 \cdot \dfrac{3}{2}t^{1/2} = \dfrac{2}{3}t^{-1/3} + \dfrac{9}{2}t^{1/2} \end{equation*}

Question 4 — Irrational and Decimal Powers

Differentiate the following function.

\begin{equation*} y = 3x^{\pi} + 6.3x^{9.1} + \pi^{9.1} \end{equation*}

Solution 4

The power rule works for any real exponent. Note that \(\pi^{9.1}\) is a constant, so its derivative is \(0\text{.}\)

\begin{equation*} \dfrac{dy}{dx} = 3\pi x^{\pi - 1} + 6.3(9.1)x^{8.1} + 0 = 3\pi x^{\pi - 1} + 57.33x^{8.1} \end{equation*}

Question 5 — Simplify, Then Differentiate

Differentiate the following function.

\begin{equation*} y = \dfrac{x + x^3 + x^2}{\sqrt{x}} \end{equation*}

Solution 5

Divide each term by \(x^{1/2}\text{:}\)

\begin{equation*} y = x^{1/2} + x^{5/2} + x^{3/2} \end{equation*}

\begin{equation*} \dfrac{dy}{dx} = \dfrac{1}{2}x^{-1/2} + \dfrac{5}{2}x^{3/2} + \dfrac{3}{2}x^{1/2} \end{equation*}

Question 6 — Distribute, Then Differentiate

Differentiate the following function.

\begin{equation*} y = \sqrt[3]{x}\,(x^2 - 3) \end{equation*}

Solution 6

Distribute \(x^{1/3}\text{:}\)

\begin{equation*} y = x^{7/3} - 3x^{1/3} \end{equation*}

\begin{equation*} \dfrac{dy}{dx} = \dfrac{7}{3}x^{4/3} - 3 \cdot \dfrac{1}{3}x^{-2/3} = \dfrac{7}{3}x^{4/3} - x^{-2/3} \end{equation*}

Question 7 — Distribute, Then Differentiate

Differentiate the following function.

\begin{equation*} y = \sqrt{x}\,(x^4 + 2) \end{equation*}

Solution 7

Distribute \(x^{1/2}\text{:}\)

\begin{equation*} y = x^{9/2} + 2x^{1/2} \end{equation*}

\begin{equation*} \dfrac{dy}{dx} = \dfrac{9}{2}x^{7/2} + 2 \cdot \dfrac{1}{2}x^{-1/2} = \dfrac{9}{2}x^{7/2} + x^{-1/2} \end{equation*}

Question 8 — Expand, Then Differentiate

Differentiate the following function.

\begin{equation*} y = (x^2 + \sqrt{x})(\sqrt{x} - x^2) \end{equation*}

Solution 8

This is a difference of squares, \((b + a)(b - a) = b^2 - a^2\text{,}\) with \(b = \sqrt{x}\) and \(a = x^2\text{:}\)

\begin{equation*} y = (\sqrt{x})^2 - (x^2)^2 = x - x^4 \end{equation*}

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

Question 9 — Tangent Line

Find an equation for the line tangent to the graph of

\begin{equation*} f(x) = 2x^3 - 2x^2 + 6 \end{equation*}

at the point \((2, 14)\text{.}\)

Solution 9

The slope of the tangent line is \(f'(2)\text{.}\) Differentiating,

\begin{equation*} f'(x) = 6x^2 - 4x \end{equation*}

\begin{equation*} f'(2) = 6(4) - 4(2) = 16 \end{equation*}

Using point-slope form through \((2, 14)\text{:}\)

\begin{equation*} y - 14 = 16(x - 2) \quad\Rightarrow\quad y = 16x - 18 \end{equation*}

Question 10 — Finding Coefficients

Determine coefficients \(a\) and \(b\) such that

\begin{equation*} p(x) = x^2 + ax + b \end{equation*}

satisfies \(p(1) = -6\) and \(p'(1) = 4\text{.}\)

Solution 10

Differentiating, \(p'(x) = 2x + a\text{,}\) so

\begin{equation*} p'(1) = 2 + a = 4 \quad\Rightarrow\quad a = 2 \end{equation*}

Then

\begin{equation*} p(1) = 1 + 2 + b = -6 \quad\Rightarrow\quad b = -9 \end{equation*}

Question 11 — Linear Approximation

Suppose that \(f(x)\) is a function with

\begin{equation*} f(135) = 28 \quad\text{and}\quad f'(135) = 3 \end{equation*}

Estimate \(f(137)\text{.}\)

Solution 11

Near \(x = 135\text{,}\) the function is approximately its tangent line:

\begin{equation*} f(x) \approx f(135) + f'(135)(x - 135) \end{equation*}

\begin{equation*} f(137) \approx 28 + 3(137 - 135) = 28 + 6 = 34 \end{equation*}

Question 12 — Interpreting a Derivative

The height of a sand dune (in centimeters) is

\begin{equation*} f(t) = 700 - 4t^2 \end{equation*}

where \(t\) is measured in years since 1995. Find \(f(4)\) and \(f'(4)\text{,}\) including units, and explain what each means in terms of the sand dune.

Solution 12

\begin{equation*} f(4) = 700 - 4(16) = 636 \text{ cm} \end{equation*}

In 1999, the dune is 636 cm tall.

\begin{equation*} f'(t) = -8t, \qquad f'(4) = -32 \text{ cm/year} \end{equation*}

In 1999, the dune’s height is decreasing at 32 cm per year.