MATH 1500 Fall 2014 Quiz 3C Solutions
Solve each of the following questions. Show all work.
[8] 1. Suppose y is a function of x defined inplicitly by
ey + x2 y 2 = csc(xy).
Determine dy/dx.
Solution: Taking derivatives of both sides gives
y dy
e
dx
dy
⇒ ey
dx
dy
⇒ ey
dx
dy
d
= − csc(xy) cot(xy) (xy)
+ 2xy + x 2y
dx
dx
dy
dy
2
2
+ 2xy + x 2y
= − csc(xy) cot(xy) y + x
dx
dx
dy
dy
+ 2xy 2 + x2 2y
= −y csc(xy) cot(xy) − x csc(xy) cot(xy)
dx
dx
2
2
Moving all the dy/dx terms to one side, factoring and then dividing yields
dy
dy
dy
+ 2x2 y
+ x csc(xy) cot(xy) = −y csc(xy) cot(xy) − 2xy 2
dx dx
dx
dy y
⇒
e + 2x2 y + x csc(xy) cot(xy) = −y csc(xy) cot(xy) − 2xy 2
dx
−y csc(xy) cot(xy) − 2xy 2
dy
= y
⇒
dx
e + 2x2 y + x csc(xy) cot(xy)
ey
3
cos(6x) + ex
[7] 2. Determine the derivative of f (x) = √
. Do not simplify.
3
x − tan x
Solution:
0
f (x) =
=
√
3
3
d √
+ ex )( 3 x − tan x) − dx
( 3 x − tan x)(cos(6x) + ex )
√
( 3 x − tan x)2
√
3
1
− 6 sin(6x) + 3x2 ex ( 3 x − tan x) − 3 x−2/3 − sec2 x (cos(6x) + e3x )
√
( 3 x − tan x)2
d
(cos(6x)
dx
[5] 3. Evaluate
tan 5x
.
x→0 sin 3x
lim
Solution:
tan 5x
sin 5x
= lim
x→0 sin 3x
x→0 cos 5x sin 3x
sin 5x
5x
5x
= lim
sin 3x
x→0
3x cos 5x
3x
sin 5x
5
5x
= lim
sin 3x
x→0
3 cos 5x
3x
5(1)
=
3(cos 0)(1)
5
= .
3
lim