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MATH 156
Instr. K. Ciesielski Spring 2002 SAMPLE TEST # 1
Solve the exercises. Show your work.
Ex. 1. Find the derivatives of the following functions. Use logarithmic differentiation where appropriate.
(a) h(x) = 2x^4 + log 3 (cosh x) (b) g(u) = arccos(eu) (c) h(t) = arctan^4 (ln t) (d) y(x) = (ln x)x^3
Ex. 2. Evaluate the integrals.
(a)
∫ (^) dx x(1 + (ln x)^2 )
(b)
∫ (^) arctan t 1 + t^2 dt
(c)
∫ (^) e 4 x 1 + e^4 x^ dx
Ex. 3. Solve for x without using a calculator.
ln x + ln(x − 1) = ln 2
Ex. 4. Evaluate the limit.
nlim→∞
( 1 − e n
)n
Ex. 5. The English language evolves in such a way that 77% of all words disappear or are replacedby noncognates every 1000 years. Of basic list of words usedby Chancer in A.D. 1400, what percentage shouldwe expect to findstill in use today?
SOLUTIONS TO EXERCISES 1–3 FROM THE SAMPLE TEST # 1
Ex. 1 (a) If h(x) = 2x^4 + log 3 (cosh x) then
h′(x) = (ln 2)2x 4 4 x^3 +
ln 3
cosh x sinh x = (4 ln 2)x^3 2 x 4
tanh x ln 3
(b) If g(u) = arccos(eu) then g′(u) = −
√^1
1 − (eu)^2
eu^ = − √ eu 1 − e^2 u^
(c) If h(t) = arctan^4 (ln t) then h′(t) = 4 arctan^3 (ln t)
1 + (ln t)^2
t
4 arctan^3 (ln t) t(1 + (ln t)^2 )
(d) If y(x) = (ln x)x^3 then ln y(x) = ln
( (ln x)x^3
) andso ln y = x^3 ln(ln x). Taking derivative from both sides and using the product rule we obtain:
1 y y′^ = 3x^2 ln(ln x) + x^3
ln x
x
Simplifying it, andmultiplying both sides by y(x) = (ln x)x^3 we obtain:
y′^ =
[ 3 x^2 ln(ln x) + x^3 x ln x
] (ln x)x 3 .
Ex. 2 (a) Let I =
∫ (^) dx x(1 + (ln x)^2 )
. Put u = ln x. Then du dx
x
. So du = dx x and
I =
1 + (ln x)^2
dx x
1 + u^2 du = arctan u + C = arctan(ln x) + C.
(b) Let I =
∫ (^) arctan t 1 + t^2 dt. Put u = arctan t. Then
du dt
1 + t^2
. So du =
dt 1 + t^2 and
I =
∫ (arctan t) dt 1 + t^2
∫ u du =
u^2 + C =
arctan^2 t + C.
(c)
∫ (^) e 4 x 1 + e^4 x^
dx =
4 e^4 x 1 + e^4 x^
dx =
ln |1 + e^4 x| + C, since numerator is the derivative of
denominator.
Ex. 3. ln x + ln(x − 1) = ln 2, so ln[x(x − 1)] = ln 2. Thus x(x − 1) = 2. Therefore x^2 − x = 2 andso ( x − 2)(x + 1) = 0. Hence we obtain that x = 2 or x = −1. But x = −1 is not a solution since, substituting to the original equation we wouldhave the expression ln(−1), which is meaningless. On the other hand x = 2 is a solution, since ln 2 + ln(2 − 1) = ln 2 is true. Answer: x = 2.
