18.01 Practice Exam 1, Solutions

1. (5 points each) Find the derivative of each of the following functions:

(a) f(x) = x^{7} + 7x; Solution: f′(x) = 7x^{6} + 7.

(b) f(x) = x/(x^{4} + 1); Solution:

f′(x) = \frac{(x^{4} + 1) − x(4x^{3})}{(x^{4} + 1)^{2}} = \frac{−3x^{4} + 1}{(x^{4} + 1)^{2}}.

(c) f(x) = 3x^{3}(x^{3} + 3)^{3}; Solution:

\begin{aligned}f′(x) & = 9x^{2}(x^{3} + 3)^{3} + 9x^{3}(x^{3} + 3)^{2}(3x^{2}) \\ & = 9[(x^{5} + 3x^{2})(x^{3} + 3)^{2} + 3x^{5}(x^{3} + 3)^{2}] = 9(4x^{5} + 3x^{2})(x^{3} + 3)^{2}). \end{aligned}

(d) f(x) = x sin x; Solution: f′(x) = sin x + x cos x.

(e) f(x) = e^{x^{3}}; Solution: f′(x) = 3x^{2}e^{x^{3}}.

(f) f(x) = cos(e^{x} + x^{1/3}); Solution: f′(x) = − sin(e^{x} + x^{1/3})(e^{x} + \frac{1}{3}x^{−2/3}).

(g) f(x) = \sqrt{x^{4} + 4};

Solution: f′(x) = \frac{1}{2}(x^{4} + 4)^{−1/2}(4x^{3}) = 2x^{3}/\sqrt{x^{4} + 4}.

(h) f(x) = x + ln(x^{2} + 1); Solution: f′(x) = 1 + \frac{2x}{x^{2} + 1}.

(i) f(x) = 2^{x}; Solution: f′(x) = 2^{x} ln 2.

(j) f(x) = cos(xe^{\sqrt{x}}). Solution:

f′(x) = sin(xe^{\sqrt{x}})\frac{d}{dx}(xe^{\sqrt{x}})
= sin(xe^{\sqrt{x}})(e^{\sqrt{x}} + xe^{\sqrt{x}} \frac{d}{dx}\sqrt{x})
= sin(xe^{\sqrt{x}})(e^{\sqrt{x}} + \frac{1}{2}\sqrt{x}e^{\sqrt{x}})
= (1 + \frac{1}{2}\sqrt{x})e^{\sqrt{x}} sin(xe^{\sqrt{x}}).

2. (5 points each) Evaluate the following limits (or explain why the limit does not exist):

(a) lim_{x→1} \frac{x^{2} − 1}{(x − 1)^{2}} ;

Solution:

lim_{x→1} \frac{x^{2} − 1}{(x − 1)^{2}} = lim_{x→1} \frac{(x + 1)(x − 1)}{(x − 1)^{2}} = lim_{x→1} \frac{x + 1}{x − 1}

does not exist, since the numerator → 2, but the denominator → 0.

(b) lim_{x→0} \frac{(x − 1) sin x}{x}.

Solution:

lim_{x→0} \frac{(x − 1) sin x}{x} = lim_{x→0}(x − 1) lim_{x→0} \frac{sin x}{x} = −1 · 1 = −1.

3. (10 points) Use the definition of the derivative to find f′(1) if f(x) = 1/x.

Solution:

f′(1) = lim_{Δx→0} \frac{f(1 + Δx) − f(1)}{Δx}
= lim_{Δx→0} \frac{1/(1 + Δx) − 1)}{Δx}
= lim_{Δx→0} \frac{−Δx/(1 + Δx)}{Δx}
= lim_{Δx→0}(−1/(1 + Δx))
= −1.

4. (10 points) An object moving on a straight line has position s = t^{3} − t^{2}.

(a) Find the velocity v and acceleration a.

Solution: v = ds/dt = 3t^{2} − 2t; a = dv/dt = 6t − 2.

(b) Which direction (positive, negative or neither) is the object moving when t = 1?

Solution: When t = 1, v = 1, so the object is moving in the positive direction.

5. (10 points) Consider the function f(x) = ^{3} \sqrt{x}.

(a) Is f(x) continuous at x = 0?

Solution: Yes. (See the graph.)

(b) Is f(x) differentiable at x = 0? If so, then find the value of the derivative. If not, then state whether the graph has a tangent line at the point (0, 0).

Solution: No, but there is a vertical tangent line (recall that ^{3} \sqrt{x} is the inverse of x^{3} which has a horizontal tangent at x = 0).

6. (10 points) Find the slope of the tangent line to the graph of y^{2} +2xy = e^{x} at the point (0, 1).

Solution:

\begin{aligned}2y \frac{dy}{dx} + 2x\frac{dy}{dx} + 2y & = & e^{x} \\ 2(x + y)\frac{dy}{dx} & = & e^{x} − 2y \\ \frac{dy}{dx} & = & \frac{e^{x} − 2y}{2(x + y)}, \end{aligned}

so the slope is −1/2.