First Quarter Calculus, Sample Exam Problems with Solutions,
Contributed by S. Cleary

For each of the questions below, choose the best answer.
NOA stands for ``None of the above.''

Question 1 If $y= -3 \csc(4x)$ then y'(0) =

a) -12
b) -48
c) 0
d) undefined
e) NOA

Answer: Using the chain rule, $y'(x) = -3(-\csc(x) \cot(x) 4) = 12 \csc(x)\cot(x)$. Since both $\csc(x)$ and $\cot(x)$ are undefined at y = 0, the answer is: d) undefined.

Question 2 If $$y = \sin^{ \displaystyle2} (4x)$$ then $y'(\displaystyle\frac{\pi}{12}) =$
a) 8
b) 0
c) $$4 \sqrt{3}$$
d) $$2 \sqrt{2}$$
e) NOA

Answer: Again, using the chain rule we find that

$$y'(x) = 2\sin(4x) {\displaystyle\frac{\displaystyle d}{\displaystyle dx}} \sin(4x) = 2\sin(4x) \cos(4x) 4,$$

or

$$y'(x) = 8 \sin(4x) \cos(4x).$$

Hence

$$y'(\displaystyle\frac{\pi}{12}) = 8\sin(4\displaystyle\frac{\pi}{12})$$

$$\cos(4\displaystyle\frac{\pi}{12}) = 8\sin(\displaystyle\frac... ...isplaystyle\frac{\pi}{3}) = 8 (\frac{1}{2})(\frac{\sqrt{3}}{2})$$

So the answer is (c): $$4 \sqrt{3}$$.

Question 3 Differentiate the following functions:
a. $y = x \cos(x)$.

Answer: Using the product rule we find

$$y'(x) = x { \displaystyle\frac{\displaystyle d}{\displaystyle... ...s(x){\displaystyle\frac{\displaystyle d}{\displaystyle dx}}(x)$$

$$y'(x) = -x \sin(x) + \cos(x).$$

b. $$y = (3x+1)^{99}$$.

Answer: Using the chain rule,

$$y'(x) = 99(3x+1)^{\displaystyle98} {\displaystyle\frac{\displ... ...e d}{\displaystyle dx} }(3x+1) = 297 (3x+1)^{\displaystyle98}.$$

c. $y = 2x^{\displaystyle2} (3x+1)$.

Answer: First multiply so that $y(x) = 6x^{\displaystyle3} + 2x^{\d s 2}$. Hence $y'(x) = 18x^{\displaystyle2} + 4x$.

d. $$y = \frac{x+2}{x+4}$$.

Answer: Using the quotient rule we find that

$$y'(x) = \frac{\displaystyle(x+4){\displaystyle d\over{\displa... ...ystyle d\over{\displaystyle dx}}( x+4)}{\displaystyle(x+4)^{2}}$$

$$y'(x) = \frac{\displaystyle x+4-x-2}{\displaystyle(x+4)^{ 2}} =\frac{\displaystyle2}{\displaystyle(x+4)^{2}}.$$

e. $$y = \frac{x^{\displaystyle3/4}}{1+x} ( 3x-3)$$

Answer: For this one, it will be easiest to first multiply out the top, then differentiate using the quotient rule. That is, $y(x) = 3\frac{\displaystyle x^{7/4} - \displaystyle x^{3/4}} {\displaystyle(1+x)}$. Hence

$$y'(x) = 3\frac{\displaystyle(1+x){\displaystyle d\over{\displ... ...{\displaystyle dx}}\displaystyle(1+x)}{\displaystyle(1+x)^{2}}.$$

Simplifying we find

$$y'(x) = \displaystyle\frac{3(3x^{\displaystyle2} + 8x - 3)}{4(x^{1/4})(1+x)^{2}}.$$

f. $$y = \sqrt{3x^{ 2} +1} = (3x^{\displaystyle2} + 1)^{\displaystyle1/2}$$.

Answer: Using the chain rule,

$$y'(x) = (1/2) (3x^{\displaystyle2} + 1)^{\displaystyle-1/2} {\displaystyle d\over{\displaystyle dx}}(3x^{\displaystyle2} + 1)$$

$$y'(x) =\frac{\displaystyle3x}{\displaystyle\sqrt{3x^{\displaystyle2} + 1}}.$$

g. $y = \displaystyle\frac{1} {\sqrt{2x+1}} = \displaystyle(2x+1)^{\displaystyle-1/2}$.

Answer: So,

$$y'(x) = (-1/2)(2x+1)^{\displaystyle-3/2)}{\displaystyle d\over{\displaystyle dx}}(\displaystyle2x+1) = \frac{-1}{(2x+1)^{3/2}}.$$

h. $y = \csc(\sin(2x)) .$

Answer: Using the chain rule we find

$$y'(x) = -\csc(\sin(2x))\cot(\sin(2x)){\displaystyle\frac{\displaystyle d} {\displaystyle dx}}(\sin(2x))$$

$$y'(x) = -\csc(\sin(2x))\cot(\sin(2x)) (\cos(2x))(2).$$

Question 4 Find the second derivative of $y=x^{\displaystyle2} \cos (3x)$

Answer: Using the product rule,

$$y'(x) = x^{\displaystyle2}{\displaystyle d\over{\displaystyle... ...3x){\displaystyle d\over{\displaystyle dx}}(x^{\displaystyle2})$$

$$= -3x^{\displaystyle2}\sin(3x) + 2x\cos(3x).$$

Using the product rule agagin (for each part) we find:

$$y''(x) = -9x^{\displaystyle2} \cos(3x) - 6x\cos(3x) - 6x\sin(3x) + 2\cos(3x).$$

Question 5 Find the first three derivatives of y = (1+2x)^4/5

Answer:

y'(x) = (2) (4/5) (1+2x)^-1/5=(8/5)(1+2x)^-1/5,

y''(x) = (2)(-1/5)(8/5)(1+2x)^-6/5 = (-16/25)(1+2x)^-6/5,

and

y'''(x) = (2)(-16/25)(-6/5)(1+2x)^-11/5 = (192/125)(1+2x)^-11/5.

Question 6 Use implicit differentiation to find the equation of the tangent line to the graph of the equation $x^{\displaystyle2} + 4 y^{\displaystyle2} = 16$ at the point $(-2, -\sqrt{3})$.

Answer: Differentiating with respect to x we obtain $2x + 8y {\displaystyle\frac{dy}{dx}} = 0$. At $(-2, -\sqrt{3})$ we have

$$2(-2) + 8(-\sqrt{3}) {\displaystyle\frac{dy}{dx}} = 0,$$

or

$${\displaystyle\frac{dy}{dx}} =-\displaystyle\frac{1}{2\sqrt{3}}.$$

Hence the equation of the line has the form

$$(y-(-2)) = -\displaystyle\frac{1}{2\sqrt{3}} (x-(-\sqrt{3})),$$

or

$$y = -\displaystyle\frac{1}{2\sqrt{3}}x + 3/2.$$

Question 7 Use implicit differentiation to find an expression for the slope of the graph of the equation $\tan(y) + y \sin (x) = 3$.

Answer: Differentiating with respect to x we obtain

$$\sec^{\displaystyle2}(y) {\displaystyle\frac{dy}{dx}} + y \cos(x) + {\displaystyle\frac{dy}{dx}}\sin(x) = 0,$$

or

$${\displaystyle\frac{dy}{dx}}(\sec^{\displaystyle2}(y) + \sin(x)) = -y\cos(x),$$

and thus,

$${\displaystyle\frac{dy}{dx}}=\displaystyle\frac{-y\cos(x)}{(\sec^{\displaystyle2}(y) + \sin(x))}.$$

Question 8 Suppose the position of an accelerating bobsled is given by $s(t) = 5 t^{\displaystyle3/2}$ feet . What is the speed of the bobsled when t = 1? What is the acceleration when t = 1?

Answer: The speed is just $s'(t)= \frac32 5 t^{\displaystyle1/2}$ so when t=1 we have that the speed is $s'(1)= \frac32 5= \frac{15}{2}$. The acceleration is given by $s''(t)= \frac32 \frac12 5 t^{\displaystyle-1/2}$ so when t=1the acceleration is $s''(1)=\frac{15}{4}$

Question 9 A little boy buys a spherical balloon of total volume 1 cubic foot. He starts blowing to fill the balloon at a rate of .1 cubic feet per minute. How fast is the radius of the balloon increasing when he has the ballon halfway blown up?

Answer: The relationship between radius and volume for a sphere of radius r is $V=\frac43 \pi r^3$, so we differentiate implicitly with respect to t to get the relationship for the rates $\frac {dV}{dt}= \frac 43 \pi 3 r^2 \frac{dr}{dt}$. The balloon is half full (one half of a cubic foot) when $\frac12=\frac43 \pi r^3$, which we solve to get $r= \sqrt[3]{\frac34 \pi}$, and we have that $\frac {dV}{dt}= .1$so we have $.1=\frac 43 \pi 3 (\sqrt[3]{\frac34 \pi})^2 \frac{dr}{dt}$ which we solve to get $\frac{dr}{dt}=.1 \frac{4}{\pi}\sqrt[3]{\frac{4}{3\pi}}$

Question 10 A cylindrical swimming pool is being filled from a fire hose at a rate of 5 cubic feet per second. If the pool is 40 feet across, how fast is the water level increasing when the pool is half full?

Answer: The volume of a cylinder is $V=\pi r^2 h$, with r=20 we get $V=\pi 400 h$, which we differentiate with respect to t to get $\frac{dV}{dt}= 400 \pi \frac{dh}{dt}$. So since $\frac{dV}{dt}=5$, we have $5=400 \pi \frac{dh}{dt}$, which gives $\frac{dh}{dt}= \frac{5}{400\pi}$.

Question 11 Implicitly differentiate $x^{\displaystyle4} + y^{\displaystyle2} + y + 3 y^{\displaystyle3} = 6$ at (1,1)

Answer: We differentiate term-by-term to get $4x^3+ 2y \frac{dy}{dx}+ \frac{dy}{dx} + 9 y^2 \frac{dy}{dx} = 0$. Substituting x=1, y=1 we get 4 + 2 y' + y' + 9 y'=0, so y'=-4/12

Question 12 Implicitly differentiate $$\sqrt{\frac{x^{\displaystyle2} + y}{x^{\displaystyle2} + 1}} =3 y$$. (Hint: it may help to simplify the equation before differentiating.)

Answer: Using the hint, we first square both sides to get $$\frac{x^{\displaystyle2} + y}{x^{\displaystyle2} + 1} =9y^2$$. We also probably want to clear out the denominator by multiplying through by x²+1 so that we can use the product rule instead of the quotient rule, which gives us x² +y=9y² x² + 9y². Differentiating both sides with respect to x gives 2x + y'= 18 y y' x² + 18 y² x + 18 y y', and solving for y' gives $$y'=\frac{-2x + 18 y^2}{1-18y x^2 -18y}$$

Question 13 On what intervals is the function $y = 3x^{\displaystyle3} - 2x$ increasing?

Answer: The function is continuous and differentiable everywhere, so we look for places where the derivative 9 x² -2 is positive, and we get 9x ²-2>0 thus x²>2/9 so $-\sqrt{2}/3>x$ and $x>\sqrt{2}/3$ are intervals with positive derivative, so the function is increasing on the intervals $(-\infty,-\frac{\sqrt{2}}{3}) \cup (\frac{\sqrt{2}}{3}, +\infty)$

Question 14 Find and classify the critical points of $$y= \frac{2x}{x^{\displaystyle2}-9}$$.

Answer: We compute the first derivative and get $$y'=\frac{2 (x^2-9)-2x (2x)}{(x^2-9)^2}$$ and then look at places where this can be zero or undefined. It can be zero when 2x²-9-4x²=0, which is not satisfied by any real x. It can be undefined when $x=\pm3$, at which point neither the function nor the first derivative are defined. So there are two critical points, which are both vertical asymptotes, at $x=\pm3$.

Question 15 Consider the function $y=x^{\displaystyle4}-3x^{\displaystyle3}$. On what intervals is it increasing ? On what intervals is it concave upwards? Find and classify the relative extrema and then use this information to sketch the graph.

Answer: The function is continuous and defined for all x. The derivative 4x³-9x² is zero only when x=0 or $x=\frac94$. So we consider the intervals $(-\infty,0)$, $(0,\frac94)$ and $(\frac94,+\infty)$. To determine the sign of the derivative on each interval we plug in convenient points, such as x=-1,1,3 and we learn that the derivative is negative (-13, -5) on the first two regions and positive (27 at x=3) on the last interval. So the function is increasing on $(-\infty,0)$, $(0,\frac94)$. For concavity, we consider the second derivative 12 x²-18x, which is zero when $x=0,\frac32$. So to understand the concavity, we look at the intervals $(-\infty,0)$, $(0,\frac32)$ and $(\frac32,+\infty)$ by plugging in conveniently-chosen points, such as x=-1,1 and 2. We see that the second derivative is negative in the middle interval and positive on the other two intervals. So the graph of the function is concave upwards on the interval $(0,\frac32)$. The candidates for relative extrema are at x=0 and $$x=\frac94$$. The first one at x=0 is not an extrema, as the function is increasing before and after x=0, so that point is neither a min nor a max. The second candidate for a relative extremum is $x=\frac94$, which is in a region where the function is concave upwards, so there is a relative minimum there. As far as plotting the function without too much work, we notice that for large postive and negative x the function is a large positive number, and we notice that it passes through a few points that are easy to plug in (0,0), (-1,4), (1,-2). So the function decreases from positive infinity down to an inflection point at the origin, where it continues to decrease but the concavity changes from upwards to downwards. The concavity switches to upwards again at $x=\frac32$ and then the function has a relative min at $x=\frac94$, after which it increases and remains concave upwards.