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Math 246 Solutions
Midterm 2
- Let f (x) =
x
. Use the limit definition to show that f ′(x) = −
x^2
Solution: Well, f ′(x) = lim h→ 0
f (x + h) − f (x) h = lim h→ 0
1 x+h −^
1 x h
= lim h→ 0
h
x + h
x
= lim h→ 0
h
x − (x + h) x(x + h)
= lim h→ 0 −h hx(x + h) = lim h→ 0
x(x + h)
x(x + 0)
x^2
- Below are some values of f, g, f ′, and g′. x 2 3 4 5 f (x) 7 3 − 2 4 g(x) 4 − 1 − 5 1 f ′(x) 11 0 9 − 3 g′(x) − 7 1 3 2
Using this information, answer the following questions:
a) If h(x) = g(x) f (x) , what is h′(5)?
Solution: Well, h′(x) =
g(x) f (x)
g′(x)f (x) − g(x)f ′(x) [f (x)]^2 , so
h′(5) = g′(5)f (5) − g(5)f ′(5) [f (5)]^2
[4]^2
b) If H(x) = g(f (x)), what is H′(5)?
Solution: Well, H′(x) =
[
g(f (x))
]′
= g′(f (x)) · f ′(x), so
H′(5) = g′(f (5)) · f ′(5) = g′(4) · f ′(5) = (3)(−3) = − 9
- Differentiate the following functions:
a) xπ^ + πx^ + eπ
Solution: Well,
xπ^ + πx^ + eπ^
= πxπ−^1 + ln(π)πx^ + 0
b) √ x 3 x + 1
Solution: Well, x √ 3 x + 1
x (3x + 1)^1 /^2
so
( x (3x + 1)^1 /^2
(x)′^
(3x + 1)^1 /^2
− (x)
(3x + 1)^1 /^2
[
(3x + 1)^1 /^2
] 2
(3x + 1)^1 /^2 − x · 12 · (3x + 1)−^1 /^2 · 3 3 x + 1
c) sin^3
7 x + 4
Solution: Well, sin^3
7 x + 4
[
sin
(7x + 4)^1 /^5
)] 3
, so
([ sin
(7x + 4)^1 /^5
)] 3 )′
[
sin
(7x + 4)^1 /^5
)] 2 (
cos
(7x + 4)^1 /^5
)) (^1
(7x + 4)−^4 /^5
d) 7cos(3x)
Solution: Well,
7 cos(3x)
= ln(7)7cos(3x)
cos(3x)
= ln(7)7cos(3x)
− sin(3x)
(3x)′^ = ln(7)7cos(3x)
− sin(3x)
- Show that
d dx
arctan(x)
1 + x^2
Solution: Let y = arctan(x), so tan(y) = x. Differentiating both sides, we get
sec^2 (y) · y′^ = 1
so y′^ =
sec^2 (y)
sec^2 (arctan(x))
Since 1 + tan^2 (θ) = sec^2 (θ) for any θ, we have 1 + tan^2 (arctan(x)) = sec^2 (arctan(x)). But tan(arctan(x)) = x, so tan^2 (arctan(x)) = x^2 , and so sec^2 (arctan(x)) = 1 + x^2. Hence,
y′^ =
sec^2 (arctan(x))
1 + x^2
- Assume that the radius r and the area A = πr^2 of a circle are differentiable functions of t.
a) Express dA dt in terms of r and dr dt
Solution: We differentiate both side with respect to t:
d dt
(A) =
d dt (πr^2 ) = π d dt (r^2 ) = π 2 r dr dt
That is, dA dt = 2πr dr dt
b) If dr dt = 7, how fast is the area changing when r = 3?
Solution: ‘How fast the area is changing’ is asking for dA dt
. We plug 7 in for dr dt and 3 in
for r into the formula for dA dt
to get
dA dt
= 2π(3)(7) = 42π
