Download Quiz 2 - Math-UMN

MATH 1271 Calculus I
October 20, 2011
Fall 2011
Lecturer: Prof. Scot Adams
Print name:
Quiz 2
Time: 30 minutes
(Show all your work for credit.)
(1) Use logarithmic differentiation to find the derivative of
√
2
f (x) = 3 x · ex +x · (x − 1)1/4 .
Solution. Recall that
f 0 (x)
d
(ln f (x)) =
.
dx
f (x)
Thus
d √
x2 +x
1/4
3
x·e
· (x − 1)
f (x) ·
ln
dx
d 1
1
2
f (x) ·
ln(x) + (x + x) ln(e) + ln(x − 1)
dx 3
4
d 1
1
2
f (x) ·
ln(x) + x + x + ln(x − 1)
dx 3
4
1
1
f (x) ·
+ 2x + 1 +
.
3x
4(x − 1)
0
f (x) =
=
=
=
1
2
(2) Evaluate the limit
lim
x→1
ln x
.
11 sin(xπ/2)(x2 − x)
(a) 1/18
(b)
1
11
(c)
1
21
(d) 1/7
(e) undefined
Solution. Naively plugging in 1 gives us an indeterminate of the form 00 . Thus we may
apply L’Hospital’s rule after first applying the product rule of limits to simplify:
ln x
1
ln(x)
= lim
· lim 2
2
x→1 11 sin(xπ/2)(x − x)
x→1 11 sin(xπ/2) x→1 x − x
(1/x)
1
= ( ) · lim
11 x→1 2x − 1
1
=
.
11
lim
(3) Differentiate
f (x) = (4 − x)sin(x) .
Solution. We use logarithmic differentiation:
d
f (x) = f (x) ·
(sin(x) ln(4 − x))
dx
sin(x)
= f (x) · cos(x) ln(4 − x) −
.
4−x
0
3
(4) Evaluate
lim (2 − x)(tan(πx/2)) .
x→1
(a) e2/π
(b) e−1
(c) e1
(d) eπ/2
(e) undefined
Proof. Plugging in 1 gives us an indeterminate of the form 1∞ . Let
L = lim (2 − x)(tan(πx/2)) .
x→1
Then
sin(πx/2)
· ln(2 − x)
x→1 cos(πx/2)
ln 2 − x
= lim sin(πx/2) · lim
x→1
x→1 cos(πx/2)
(−1/(2 − x))
= (1) · lim
x→1 (π/2) sin(πx/2)
1
=
2/π
2
.
=
π
Thus as ln L = π2 , it follows that L = eπ/2 .
ln(L) = lim