MATH 133, section 1 – Calculus With Fundamentals 1 Exam 3 Practice Problems November 4, 2015

I. Do not use the differentiation rules from Chapter 3 in this question.

A) State the limit definition of the derivative f′(x).

Solution: f′(x) = lim_{h→0} \frac{f(x+h)−f(x)}{h}. B) Use the definition to compute the derivative function of f(x) = \frac{1}{3x}.

Solution:

f′(x) = lim_{h→0} \frac{\frac{1}{3(x+h)} − \frac{1}{3x}}{h}

= lim_{h→0} \frac{3x − (3x + 3h)}{9hx(x + h)}

= lim_{h→0} \frac{−3h}{9hx(x + h)}

= lim_{h→0} \frac{−3}{9x(x + h)}

= \frac{−3}{9x^{2}} = \frac{−1}{3x^{2}} .

(Note: this agrees with the result of applying the chain rule to f(x) = (3x)^{−1}.) C) Find the equation of the line tangent to the graph y = \frac{1}{3x} at x = 2.

Solution: At x = 2, f(2) = \frac{1}{6} and f′(2) = \frac{−1}{12}. So the equation of the tangent line is y − \frac{1}{6} = (\frac{−1}{12}) (x − 2), or y = \frac{−1}{12}x + \frac{1}{3}.

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.

A)

\frac{d}{dx} (5x\sqrt{x} − \frac{2}{x^{3}} + 11x − 4)

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′ = \frac{15}{2}x^{1/2} + 6x^{−4} + 11.

B)

\frac{d}{dt} (\frac{t^{2}e^{3t}}{t^{4} + 1})

Solution: By the quotient rule, product rule, and chain rule the derivative is:

\frac{(t^{4} + 1) · (3t^{2}e^{3t} + 2te^{3t}) − (t^{2}e^{3t}) · (4t^{3})}{(t^{4} + 1)^{2}}.

\frac{d}{dz} \frac{z^{2} − 2z + 4}{z^{2} + 1}

Solution: Again using the quotient rule, the derivative is:

\frac{(z^{2} + 1)(2z − 2) − (z^{2} − 2z + 4)(2z)}{(z^{2} + 1)^{2}} = \frac{2z^{2} − 6z − 2}{(z^{2} + 1)^{2}}.

D)

\frac{d}{dx} (\frac{π^{2} + tan(e^{π}) − 2x^{e}}{4} )

Solution: Most of this is constant; only the x^{e} from the last term on the top contributes anything nonzero to the derivative:

\frac{−e}{2}x^{e−1}.

E)

\frac{d}{dx} (sin(x) (x^{7} − \frac{4}{\sqrt{x}}))

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).

F) Find y′ (note this is just another way of asking the same question!)

y = (e^{2x} + 2)^{3}

Solution: By the chain rule, the derivative is:

3(e^{2x} + 2)^{2}(2e^{2x}) = 6e^{2x}(e^{2x} + 2)^{3}.

G) Find y′

y = \frac{x + 1}{3x^{4} − 1}

Solution: By the quotient rule,

y′ = \frac{(3x^{4} − 1)(1) − (x + 1)(12x^{3})}{(3x^{4} − 1)^{2}} = \frac{−9x^{4} − 12x^{3} − 1}{(3x^{4} − 1)^{2}}.

H) Find y′

y = \frac{sin(x)}{1 + cos(x)} + x^{2} cos(x^{3} + 3)

Solution: By the quotient, product, and chain rules:

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

III. The total cost (in $) of repaying a car loan at interest rate of r% per year is C = f(r).

A) What is the meaning of the statement f(7) = 20000? Solution: At an interest rate of 7% per year, the cost of repaying the loan is 20000 dollars.

B) What is the meaning of the statement f′(7) = 3000? What are the units of f′(7)? 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).

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 ≥ 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.)

(a) Over which intervals with t ≥ 0 is the amount increasing? (i.e. Q′(t) > 0) decreasing (i.e. Q′(t) < 0)? Solution: Q′(t) = 3t^{2} − 6t + 1. Q′(t) = 0 when

t = \frac{6 ± \sqrt{36 − 12}}{6} = 1 ± \frac{\sqrt{6}}{3} ≐ 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).

(b) Over which intervals is the rate of change of Q increasing? decreasing? 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).

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 = \frac{4πr^{3}}{3} and the surface area is A = 4πr^{2}.)

Solution: By the chain rule, \frac{dV}{dt} = 4πr^{2} \frac{dr}{dt}. We are given \frac{dV}{dt} = 20 when r = 6, so

\frac{dr}{dt} = \frac{20}{4π(6)^{2}} = \frac{5}{36π}

inches per minute. The rate of change of the surface area is

\frac{dA}{dt} = 8πr \frac{dr}{dt} = \frac{48π · 5}{36π} = \frac{20}{3}

square inches per minute. VI. Review problems 2 and 3 from Quiz 5.