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\centerline{MATH 133, section 1 -- Calculus With Fundamentals 1}
\centerline{Exam 3 Practice Problems}
\centerline{November 4, 2015}
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\noindent
I. Do not use the differentiation rules from Chapter~3 in this
  question.
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\item{A)} State the limit definition of the
  derivative~$f'(x)$.
\item{} {\it Solution:}  $f'(x) = \lim_{h\to 0} {f(x+h) - f(x)\over h}$.
\item{B)} Use the definition to
    compute the derivative function of $f(x) = {1\over 3x}$.
\item{} {\it Solution:}
$$\eqalign{f'(x) &= \lim_{h\to 0} {{1\over 3(x+h)} - {1\over 3x}\over h}\cr
&= \lim_{h\to 0} {3x - (3x + 3h)\over 9hx(x+h)}\cr
&= \lim_{h\to 0} {-3h\over 9hx(x+h)}\cr
&= \lim_{h\to 0} {-3\over 9x(x+h)}\cr
&= {-3\over 9x^2} = {-1\over 3x^2}.}$$
(Note: this agrees with the result of applying the chain rule to $f(x) = (3x)^{-1}$.)
\item{C)} Find the equation of the line tangent to
    the graph $y = {1\over 3x}$ at $x=2$.
\item{}  {\it Solution:}  At $x = 2$, $f(2) = {1\over 6}$ and 
$f'(2) = {-1\over 12}$.  So the equation 
of the tangent line is $y - {1\over 6} = \left({-1\over 12}\right)(x - 2)$, 
or $y = {-1\over 12}x + {1\over 3}$.  
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\noindent
II.  Use the sum, product, quotient, and/or chain  
  rules to compute the following derivatives. You may use any correct
  method, but must show work for full
  credit.
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\item{A)}
$${d\over dx}\left(5x\sqrt{x}-{2 \over x^3}+ 11x- 4\right)$$
\item{} {\it Solution:}  The function can also be written as $5x^{3/2} - 2x^{-3} + 11x - 3$.
In this form, we only need the power rule to differentiate:
$$y' = {15\over 2} x^{1/2} + 6x^{-4} + 11.$$
\item{B)}
$${d\over dt}\left({t^2e^{3t}\over t^4 + 1}\right)$$
\item{}  {\it Solution:}  By the quotient rule, product rule, and chain rule the derivative is:
$${(t^4 + 1)\cdot (3t^2e^{3t} + 2te^{3t}) - (t^2e^{3t})\cdot (4t^3)\over (t^4 + 1)^2}.$$
\item{C)}
$${d\over dz} {z^2 - 2z + 4\over z^2 + 1}$$
\item{} {\it Solution:}   Again using the quotient rule, the derivative is:
$${(z^2 + 1)(2z - 2) - (z^2 - 2z + 4)(2z)\over (z^2 + 1)^2} = 
{2z^2 - 6z - 2\over (z^2 + 1)^2}.$$
\item{D)}
$${d\over dx}\left({\pi^2+\tan(e^\pi)-2x^e\over 4}\right)$$
\item{} {\it Solution:}  Most of this is constant; only the $x^e$ from the last
term on the top contributes anything nonzero to the derivative:
$${-e\over 2}x^{e-1}.$$
\item{E)} 
$${d\over dx}\left( \sin(x)\left(x^7 - {4\over \sqrt{x}}\right)\right)$$
\item{} {\it Solution:}   Rewrite the fuction as $\sin(x)(x^7 - 4x^{-1/2})$.  Then by the 
product rule the derivative is:
$$\sin(x)(7x^6 + 2x^{-3/2}) + (x^7 - 4x^{-1/2})\cos(x).$$
\item{F)} Find $y'$ (note this is just another way of asking the same question!)
$$y = (e^{2x} + 2)^3$$
\item{} {\it Solution:}  By the chain rule, the derivative is: 
$$3(e^{2x} + 2)^2(2e^{2x}) = 6e^{2x}(e^{2x} + 2)^3.$$
\item{G)}  Find $y'$  
$$y = {x + 1\over 3x^4 - 1}$$
\item{} {\it Solution}: By the quotient rule,
$$y' = {(3x^4 - 1)(1) - (x+1)(12x^3) \over (3x^4-1)^2} = {-9x^4-12x^3-1 \over (3x^4-1)^2}.$$
\item{H)}  Find $y'$
$$y = {\sin(x) \over 1 + \cos(x)} + x^2 \cos(x^3 + 3)$$
\item{} {\it Solution:}  By the quotient, product, and chain rules:
$$\eqalign{y' &= {(1 + \cos(x))\cos(x) - \sin(x)(-\sin(x))\over (1 + \cos(x))^2} - x^2\sin(x^3+3)(3x^2) + 2x \cos(x^3 + 3)\cr
&= {1 + \cos(x) \over (1 + \cos(x))^2} - 3x^4\sin(x^3 + 3) + 2x \cos(x^3 + 3)\cr
&= {1\over 1 + \cos(x)} - 3x^4 \sin(x^3+3)+ 2x \cos(x^3 + 3).}$$
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\noindent
III. The total cost (in $ \$ $) of repaying a car loan at 
interest rate of $r \%$ per year is $C = f(r)$.  
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\item{A)} What is the meaning of the statement $f(7) = 20000$?
\item{} {\it Solution:}  At an interest rate of $7\%$ per year, the cost of repaying
the loan is $20000$ dollars.  
\item{B)} What is the meaning of the statement $f'(7) = 3000$? What are the units of $f'(7)$? 
\item{} {\it Solution:}  At an interest rate of $7\%$ per year, the rate of change
of the cost of repaying the loan is $3000$ dollars per ($\%$ per year).  
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\noindent
IV.  The quantity of a reagent present in a chemical 
reaction is given by $Q(t) = t^3 - 3t^2 + t + 30$ grams at time $t$ seconds for all $t \ge 0$.
(Note:  For a question like this, I could also give you the plot of the function and ask
questions like those below.  In this case you need to start from the formula and compute
$Q'(t)$; if you were given the graph, you need to make the connection between slopes of 
tangent lines and signs of $Q'(t)$ visually.)
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\item{(a)}  Over which intervals with $t \ge 0$ is the amount increasing? (i.e. $Q'(t) > 0$) decreasing (i.e. $Q'(t) < 0$)?
\item{} {\it Solution:}  $Q'(t) = 3t^2 - 6t + 1$.  $Q'(t) = 0$ when 
$$t = {6 \pm \sqrt{36-12}\over 6} = 1\pm {\sqrt{6} \over 3} \doteq 1.816, .184.$$
Since this is a quadratic function with a positive $t^2$ coefficient, $Q'(t) > 0$
for $t > 1.816$ and $t < .184$.  $Q'(t) < 0$ for $.184 < t < 1.816$ ($t$ in seconds).
\item{(b)}  Over which intervals is the rate of change of $Q$ increasing? decreasing?
\item{} {\it Solution:}   The rate of change of $Q$ is increasing when $(Q')' > 0$ and decreasing
when $(Q')' < 0$.  The second derivative of $Q$ is $Q''(t) = 6t - 6$.  So 
$Q''(t) > 0$ for $t > 1$ and $Q''(t) < 0$ for $t < 1$ ($t$ in seconds).
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\noindent
V.  A spherical balloon is being inflated at 20 cubic inches per
minute.  When the radius is $6$ inches, at what rate is the 
radius of the balloon increasing?  At what rate is the surface
area increasing?  (The volume of a
sphere of radius $r$ is $V = {4\pi r^3\over 3}$ and the surface
area is $A = 4\pi r^2$.)
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\noindent
{\it Solution:}  By the chain rule, ${dV\over dt} = 4\pi r^2 {dr \over dt}$.
We are given ${dV\over dt} = 20$ when $r = 6$, so 
$${dr\over dt} = {20 \over 4\pi (6)^2} = {5\over 36\pi}$$ inches per minute.
The rate of change of the surface area is
$${dA\over dt} = 8\pi r {dr\over dt} = {48\pi \cdot 5\over 36\pi} = {20\over 3}$$
square inches per minute.
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\noindent
VI.  Review problems 2 and 3 from Quiz 5.  
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