Download ANSWERS 1. d sin(x2)=2xcos(x2). 2. First derivative: d tan3(ex) = 3

Calculus I, Fall 08
Grinshpan
ANSWERS
1.
d
dx
sin(x2 ) = 2x cos(x2 ).
d
2. First derivative: dx
tan3 (ex ) = 3 tan2 (ex ) sec2 (ex )ex .
Second derivative using log differentiation:
d2
d sin2 (ex ) x
3 x
e
tan (e ) = 3
dx2
dx cos4 (ex )
sin2 (ex ) x
e (2ex cot(ex ) + 4ex tan(ex ) + 1) .
=3
cos4 (ex )
3. Let x = x(t) be the distance from the bottom end to the wall and
let θ = θ(t) be the angle in question. Since x = 10 cos
√θ, we have√
ẋ = −10 sin θ θ̇. When x = 2, ẋ = −1/2 and sin θ = 96/10 = 52 6.
√
Hence θ̇x=2 = 1/8 6 (rad/s).
4. Write tan3 (xy 2 + y) = x as tan(xy 2 + y) = x1/3 . Then
xy 2 + y = arctan(x1/3 )
x−2/3
1 + x2/3
1/3
(2xy + 1)y 0 + y 2 = 2/3
x + x4/3
13
2
2/3
4/3 − y
y 0 = x +x
2xy + 1
y 2 + x2yy 0 + y 0 =
1
3
5.
x3 + y 3 = 1
3x2 + 3y 2 y 0 = 0
x2 + y 2 y 0 = 0
(y 0 = −x2 /y 2 )
2x + 2y(y 0 )2 + y 2 y 00 = 0
x4
+ y 2 y 00 = 0
4
y
2 00
y y = −2x(1 + x3 /y 3 ) = −2x/y 3
2x + 2y
y 00 = −2x/y 5 .
2
√
√
6. Check that (1, 4 15) is on the curve: 14 + ( 4 15)4 = 1 + 15 = 16.
Compute the slope:
x4 + y 4 = 16
4x3 + 4y 3 y 0 = 0
y 0 = −(x/y)3
y0 √
= −15−3/4 .
4
(1,
15)
Equation of the tangent line: y = − 1513/4 (x − 1) + 151/4 .
7.
√
3
y = x x2 + 1
1
ln y = ln x + ln(x2 + 1)
3
1
1 2x
(ln y)0 = +
2
x 3 x + 1 √
1 2
1 1 2x
x
3
0
2
+
+
y =y·
=x x +1
.
x 3 x2 + 1
x 3 x2 + 1
8.
1
d
log2 x =
dx
x ln 2
d
d ln 2
ln 2
.
logx 2 =
=−
dx
dx ln x
x ln2 x
9.
d
(1/2)x = (1/2)x ln(1/2) = − ln 2 · 2−x .
dx
√
x
d 1/x
2
2 = 21/x ln 2 · (−1/x2 ) = − ln 2 2 .
dx
x
10.
d x
d
x =
(ex ln x ) = ex ln x (ln x + 1) = xx (ln x + 1).
dx
dx
11.
d
1
1/(x2 + 1)
=−
.
dx arctan x
arctan2 (x)
3
12.
e2x arcsin(−x)
x cos2 x
ln y = 2x + ln(arcsin(−x)) − ln x − 2 ln cos x
√
1/ 1 − x2 1
0
(ln y) = 2 +
− + 2 tan x
arcsin
x
x
√
2
1/ 1 − x
1
0
y =y· 2+
− + 2 tan x
arcsin x
x
√
1/ 1 − x2 1
e2x arcsin(−x)
0
2+
− + 2 tan x
y =
x cos2 x
arcsin x
x
y=