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Handout Exam 2 - Spring 2026

Section Topics

1.8-2.7
Be sure to try each question before looking at the solutions.

Exercises Questions

1.

Two differentiable functions \(u\) and \(v\) are graphed below near \(x=1\) . Note: These graphs are symmetric about a horizontal line defined where they intersect.
Figure 230. The graphs of \(u(x)\) and \(v(x)\text{.}\)
  1. Without formulas, determine the sign (positive, negative, or zero) of \((uv)'(1)\) . Explain using the product rule and the picture.
  2. Determine the sign of \(\left(\dfrac{u}{v}\right)'(1)\) . Explain using the quotient rule and the picture.
Answer.
  1. At \(x=1\) , the graph shows: \(u(1) > 0, u'(1) > 0,\) \(v(1) > 0,{\text{ and }}v'(1) = - u'(1) < 0.\) Note that
    \begin{equation*} (uv)'(1) = u'(1)v(1)+u(1)v'(1). \end{equation*}
    The first term is positive (positive slope times positive value). The second term is negative (positive times negative). Since, \(v'(1)=-u'(1)\) and \(0 < u(1) < v(1)\) , the first (positive) term dominates.
  2. Note that
    \begin{equation*} \displaystyle \left( \frac{u}{v}\right)'(1) = \frac{u'(1)v(1)-u(1)v'(1)}{v(1)^{2}}. \end{equation*}
    The denominator is positive. In the numerator, \(u'(1)v(1) > 0\) and \(-u(1)v'(1)\) is also positive since \(v'(1)<0\) . Adding these is positive. Hence the fraction is also positive.

2.

The function \(f\) is strictly increasing and differentiable. Its graph and a tangent line at \(x=b\) are shown in Figure 231. The slope of the tangent line is \(\displaystyle m=\frac{3}{2}\) . Let \(g=f^{-1}\) . Compute \(g'\big(f(b)\big)\) .
Figure 231. Graph a function \(f(x)\) and its tangent line at \(x=b\text{.}\)
Answer.
Slope of tangent to \(f\) at \(x=b\) is \(m = \frac{3}{2}.\) Since \(g=f^{-1}\) ,
\begin{equation*} \displaystyle g'(f(b)) = \frac{1}{f'(b)}= \frac{1}{3/2}= \frac{2}{3}. \end{equation*}

3.

Let \(F(x)=\sqrt{4x^{2}+3}\) . Compute \(F'(x)\) and then evaluate \(F'(2)\) (exact value only).
Answer.
\begin{align*} F'(x) \mathstrut \amp =\frac{1}{2}(4x^{2}+3)^{-1/2}\cdot 8x \\ \amp = \frac{4x}{\sqrt{4x^{2}+3}}. \end{align*}
so that
\begin{align*} F'(2)\mathstrut \amp =\frac{8}{\sqrt{16+3}} \\ \amp = \frac{8}{\sqrt{19}}. \end{align*}

4.

A function \(f\) is differentiable and satisfies \(f(3)=7\) and \(f'(3)=-1.6\) .
  1. Find the linearization \(L(x)\) at \(x=3\) .
  2. Use \(L(x)\) to estimate \(f(3.3)\) .
  3. If you are told \(f''(3)>0\) , explain in one sentence whether your estimate in (b) is likely an overestimate or underestimate of \(f(3.3)\) .
Answer.
  1. \(\displaystyle L(x)=f(3)+f'(3)(x-3)=7-1.6(x-3).\)
  2. \(\displaystyle f(3.3)\approx L(3.3)=7 -1.6(0.3)=7-0.48 = 6.52.\)
  3. If \(f''(3)>0\) , \(f\) is likely concave up near \(x=3\) . The tangent line then lies below the curve, so the linear approximation is an underestimate.

5.

Compute each derivative.
  1. \(\displaystyle h(x)= 5x^{4} - 3x + \dfrac{2}{x}\)
  2. \(\displaystyle q(x)= (3x^{2}+1)\sin x\)
  3. \(\displaystyle r(x)= \dfrac{x^3}{2x-5}\)
Answer.
  1. \(\displaystyle h'(x)=20x^{3} - 3 - \frac{2}{x^{2}}.\)
  2. \(\displaystyle q'(x)=6x\sin x + (3x^{2}+1)\cos x.\)
  3. Use the quotient rule:
    \begin{align*} r'(x) \mathstrut \amp =\frac{3x^{2}(2x-5)-x^{3}(2)}{(2x-5)^{2}}\\ \amp =\frac{6x^{3}-15x^{2}-2x^{3}}{(2x-5)^{2}}\\ \amp =\frac{4x^{3}-15x^{2}}{(2x-5)^{2}}. \end{align*}

6.

The height \(H(t)\) (in meters) of water in a tank is decreasing at a rate of \(0.12\) m/min. The water temperature \(T\) (in \(^{\circ}\) C) changes with height according to \(T(H)=18 + 0.5H.\)
  1. Compute \(\dfrac{dT}{dt}\) at the instant the water is 2 m deep.
  2. Interpret your answer with units in one sentence.
Answer.
  1. From the chain rule,
    \begin{align*} \frac{dT}{dt}\mathstrut \amp =\frac{dT}{dH}\frac{dH}{dt}\\ \amp =0.5(-0.12) \\ \amp = -0.06\ \text{°C/min}. \end{align*}
  2. The water temperature is decreasing at a rate of \(0.06^{\circ}\text{C per minute}.\)

7.

Find the equation of the tangent line to the curve \(\displaystyle x^{2}+xy+4y^{2}=10\) at the ordered pair \((2,1)\) .
The graph of a given ellipse and the ordered pair at (2,1).
Figure 232. A graph of \(\displaystyle x^{2}+xy+4y^{2}=10 \text{.}\)
Answer.
Differentiate:
\begin{equation*} 2x + (x y' + y) + 8y y' = 0. \end{equation*}
Group \(y'\) terms:
\begin{equation*} y'(x+8y)=-(2x+y), \end{equation*}
\begin{equation*} y' = \frac{-(2x+y)}{x+8y}. \end{equation*}
At \((2,1)\) :
\begin{equation*} y'=\frac{-(4+1)}{2+8}=\frac{-5}{10}=-\frac{1}{2}. \end{equation*}
Tangent line:
\begin{equation*} y-1 = -\frac{1}{2}(x-2). \end{equation*}

8.

You know that at \(x=4\text{,}\)
\begin{equation*} u(4)=3,\quad u'(4)=2.1,\quad v(4)=5,\quad v'(4)=-0.6. \end{equation*}
Let \(P(x)=u(x)v(x)\) and \(Q(x)=\dfrac{u(x)}{v(x)}\text{.}\)
  1. Compute \(P'(4)\) .
  2. Compute \(Q'(4)\) .
  3. Explain in one sentence whether \(P\) is increasing or decreasing at \(x=4\) .
Answer.
  1. We have
    \begin{align*} P'(4)\mathstrut \amp =u'(4)v(4)+u(4)v'(4) \\ \amp =2.1(5)+3(-0.6) \\ \amp =10.5 - 1.8 \\ \amp = 8.7. \end{align*}
  2. We have
    \begin{align*} Q'(4) \mathstrut \amp =\frac{u'(4)v(4)-u(4)v'(4)}{v(4)^{2}}\\ \amp =\frac{2.1(5)-3(-0.6)}{25} \\ \amp =\frac{10.5+1.8}{25} \\ \amp =\frac{12.3}{25} \\ \amp = 0.492. \end{align*}
  3. Since \(P'(4)=8.7>0\) , \(P \text{ is increasing at }x=4.\)

9.

The graph of a periodic function \(p(x)\) is shown. Which of the following is most likely to represent \(p'(x)\) ? [Check the appropriate box(es).]
The graph of a periodic function.
Figure 233. A graph of \(p(x)\) which is periodic.
\(\square\)
\(\displaystyle \sin(2x)\)
\(\square\)
\(\displaystyle -\cos(2x)\)
\(\square\)
\(\displaystyle 2\sin(x)\)
\(\square\)
\(\displaystyle -2\cos(x)\)
\(\square\)
\(\displaystyle 2\sin(2x)\)
\(\square\)
\(\displaystyle -2\cos(2x)\)
\(\square\)
\(\displaystyle 2\sin(x)-2\)
\(\square\)
\(\displaystyle -2\cos(x)-2\)
\(\square\)
\(\displaystyle -2\sin(x)\)
\(\square\)
\(\displaystyle 2\cos(x)\)
\(\square\)
\(\displaystyle 2\sin(\frac{1}{2}x)\)
\(\square\)
\(\displaystyle -2\cos(\frac{1}{2}x)\)
Answer.
\(2\sin(2x)\)

10.

A function \(f\) and its derivative \(f'\) take the values given in Table 234. If \(p(x)=f(2e^{x})\text{,}\) compute \(p'(0)\text{.}\)
Table 234. Values of \(f\) and \(f'\) at various values.
\(x\) 0 1 2
\(f(x)\) 2 1 0
\(f'(x)\) 1 0 2
Answer.
Note that
\begin{equation*} p'(x)=f'(2e^{x})\cdot 2e^{x}. \end{equation*}
At \(x=0\) :
\begin{equation*} p'(0)=f'(2e^{0})\cdot 2e^{0} = f'(2)\cdot 2. \end{equation*}
From the table: \(f'(2)=2\) . So \(p'(0)=4.\)

11.

A company sells a product at price \(p\) dollars. The number of units sold \(q(p)\) depends on the price. The revenue is \(R(p) = p \cdot q(p)\) . When the price is \(p = \$20\) , the number of units sold is \(q(20)=600\) . It is also known that \(q'(20)=-18\) .
  1. In this context, what does \(q'(20)=-18\) mean? Use units in giving your one-sentence explanation.
  2. Compute \(R'(20)\) .
  3. Interpret the meaning of the sign of \(R'(20)\) .
  4. Assuming that the company wishes to maximize revenue, should the company raise or lower the price at \(p=\$20\) ? Explain briefly.
Answer.
  1. \(q'(20)=-18\) means that for each $1 increase in price at \(p=\$20\) , unit sales decrease by 18 units (units per dollar).
  2. \(R'(p)=q(p)+p q'(p),\) so that \(R'(20)=600+20(-18)=600-360= 240.\)
  3. Since \(R'(20)>0\) , revenue is increasing at \(p=20\) . This means a small price increase will raise revenue.
  4. To maximize revenue, since \(R'(20)>0\) , the company should raise the price (slightly).
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