Sample Midterm Exam 2 Solutions

Question 1 (10). Find the derivative of the following functions:
a. $f(x) = x^8 \sin 5x$
Solution:
$f'(x) = 8x^7 \sin 5x + x^8 (\cos 5x) 5$ by the product rule and the chain rule.

b. $f(x) = (x + \sin x)^{23}$
Solution:
$f'(x) = 23(x + \sin x)^{22}(1 + \cos x)$
by the chain rule.

c. $f(x) = \frac{x^2 - x}{ \tan x}$
Solution:
$f'(x) = \frac{(\tan x)(2x - 1) - (\sec ^2 x) (x^2 - x)}{ \tan ^2 x}$
by the quotient rule.

d. $f(x) = \frac{1 + (1/x)}{1 - (1/x)}$
Solution:
First simplify: this gives
$f(x) = \frac{x + 1}{x -1}$
Now it is easy to use the quotient rule:
$f'(x) = \frac{(x - 1)(1) - (x+1) (1)}{(x-1)^2}$
by the quotient rule. This can be simplfied to $\frac{-2}{(x-1)^2}$

e. $f(x) = (2x+1)^7 (3x+1)^5$
Solution:
$f'(x) = 7(2x+1)^6(2) (3x+1)^5 + 5(3x+1)^4(3) (2x+1)^7$
by the product and chain rules.

Question 2 (15). a. State the definition of the derivative of a function $f$ at a point $x$.

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)} {h}$, if this limit exists.

b. Use the definition of the derivative to compute $f'(x)$ for $f(x) = \frac{2}{x}$.
$\frac{f(x+h) - f(x)} {h} = \frac{\frac{2}{x+h} - \frac{2}{x}} {h} \\ = \frac{-2} {(x+h)x}$
Taking the limit of this as $h \to 0$, we obtain $fÕ(x) = \frac{-2}{x^2}$.

Question 3 (10). Find all the vertical and horizontal asymptotes of the graph of $f(x) = \frac{x^2 - 2x + 1}{x^2 - 1}$
We first notice that there is a simplification,
$f(x) = \frac{x - 1}{x + 1}$
Then we see that there is a vertical asymptote at $x=-1$. As $x \to \infty$, we have $f(x) \to 1$, and similarly for $x\to -\infty$. So there is one horizontal asymptote, $y=1$, for both $x \to \infty$ and $x\to -\infty$.

Question 4 (20). For each of the following, either find the limit or state that "no limit exists" and briefly explain why. Show work used to get your answer.

a. $\lim_{x \to 0} \frac{2 + 3\sin x} {x^3 + 1}$
Answer: 2. The limit can be obtained by plugging in, since this does not lead to division by zero or other problems, and the numerator and denominator are continuous.

b. $\lim_{x \to \infty} \cos x$.

Does not exist, since $\cos x$ oscillates between -1 and 1 and does not approach a single value.

c. $\lim_{x \to \infty} \frac{\cos x}{x^2 - \sin x}$.
Answer: 0, since $\vert\frac{\cos x}{x^2 - \sin x} \vert \le \frac{1}{x^2 - 1}$ for $x$ large, and this approaches zero as $x \to \infty$.

d. $\lim_{x \to 2} \frac {x^2 - 4} { x-2}$.

Answer: 4, since $\frac {x^2 - 4} { x-2} = (x+2)$ when $x \neq 2$, and $\lim_{x \to 2} { x+2} = 4.$

Question 5 (10). For the function $f(x) = x^2 + 2 \tan x - 2$

a. Find the equation of the tangent line to the graph of $f(x)$ at the point (0,-2).
$f'(x) = 2x + 2\sec ^2 (x)$ and at $x=0$ this has value $f'(0) =2\sec ^2 (0) = 2$. So the line has slope 2 and goes through the point (0,-2). Using the point-sl ope formula gives the equation
$y - (-2) = 2 (x - 0)$ or $y = 2x - 2$.

b. Show that $f(x) = 0$ at some point.
We know that $f(0) = -2$ and $f(\pi/4) = (\pi/4)^2 + 2 - 2 = (\pi/4)^2 > 0$. The function $f(x)$ is continuous on the interval $[0,\pi/4]$, so by the Intermediate Value Theorem, there is a point in this interval where $f(x) = 0.$

Question 6 (10) a. State the precise definition of what is meant by $\lim_{x \to a} f(x) = L$.
Given an $\epsilon > 0$ there is a $\delta$ such that whenever $0 < \vert x-a\vert < \delta$ then it is true that $\vert f(x) - L\vert < \epsilon$.

Use the precise definition of the limit to prove that $\lim_{x \to 0} 5 x^2 - 4 = -4$.
To ensure that $\vert f(x) -4\vert < \epsilon$, or $\vert(5 x^2 - 4 ) -4\vert < \epsilon$, what $x$ can we allow? Simplifying gives $\vert 5x^2\vert < \epsilon$, or $\vert x\vert < \sqrt {\frac {\epsilon}{5}}$. So we pick $\delta = \sqrt {\frac {\epsilon}{5}}$ and we satisfy the condition for the limit to equal 4.

Question 7 (5) Give an example of a function $f(x)$ which is continuous at $x=1$ but not differentiable at $x=1$.

The function $f(x) = \vert x-1\vert$ is an example.

Question 8 (5) Suppose $f$ and $g$ are functions and $f(3) = 2,\ f'(3) = 4,\ g(5) = 3,\ g'(5) = 7.$ Where can you calculate the derivative of $f \circ g$? What is it equal to?
At $x=5$, the chain rule tells us that $(f \circ g)'(5) = (f'(g(5))(g'(5) = f'(3)(7) = (4)(7) = 28$

Question 9 (5) Let $f(x) = \sqrt[4] {x^5}$. Find $f'(16)$.
By the chain rule, $f'(x) = (5/4) x^ {1/4}$ and $f'(16) = (5/4)((16)^ {1/4} = 5/2.$

Question 10 (10). Find an anti-derivative of the following functions:
a. $f(x) = 5/x^2$.
$F(x) = -5x^{-1}$.

b. $f(x) = 3 \sin x$
$F(x) = -3 \cos x$.

c. $f(x) = \csc x \cot x$
$F(x) = - \csc x$.