2.1 Homework Solutions
1. f (x) = x5 + 6x 2 − 7x
f ′ ( x ) = 5x 4 +12x − 7
f ′′ ( x ) = 20x 3 +12
(
)
2
y′ = ( x +1)
3
3. y = x 3 + 1
2
3
3
( x +1)
y′′ =
3
=
1
− 13
3
⋅ 3x 2 =
3
(
1
3
)
(
) ⎡⎣⎢( x +1) ⋅ 4x − 2x
( x +1)
−
3
⋅ 4x − 2x ⋅ 1 3 x 3 +1 3 ⋅ 3x 2 x +1
=
2
3
3
x +1
4x 4 + 4x − 2x 4
( x +1)
)
x 3 +1
2
(
4
(
2x 2
3
=
(
)
2x x 3 + 2
( x +1)
3
4
2
)
−2 3
3
3
2
3
3
5. g(t) = t 3e5t
g′ (t ) = t 3 ⋅e5t ⋅5 + e5t ⋅ 3t 2 = t 2 e5t ⎡⎣ 5t + 3⎤⎦
g′′ (t ) = t 2 e5t ⋅5 + t 2 ( 5t + 3) ⋅e5t ⋅5 + e5t ⋅ ( 5t + 3) ⋅2t
= te5t ⎡⎣ 5t + 5t ( 5t + 3) + 2 ( 5t + 3) ⎤⎦
= te5t ⎡⎣ 25t 2 + 30t + 6 ⎤⎦
7. y = sin 3 x
y′ = 3sin 2 x cos x
y′′ = 3 ⎡⎣sin 2 x ⋅−sin x + cos x ⋅2sin x cos x ⎤⎦ = 3sin x ⎡⎣ 2cos2 x − sin 2 x ⎤⎦
1 4⎤
⎦⎥
2.2 Homework Solutions
1.
sec y = x 2 y + ex
dy
dy
= x 2 + y ( 2x ) + ex
dx
dx
dy
sec y tan y − x 2
= 2xy + ex
dx
x
dy
2xy + e
=
dx sec y tan y − x 2
sec y tan y
(
3.
ysin x +
)
x
= 16y
sin y
dy
dy sin y (1) − x cos y dx
dy
ycos x + sin x +
= 16
2
dx
dx
sin y
dy
dy
dy
ycos xsin 2 y + sin xsin 2 y + sin y − x cos y = 16sin 2 y
dx
dx
dx
dy
ycos xsin 2 y + sin y = x cos y − sin xsin 2 y16sin 2 y
dx
2
dy
ycos xsin y + sin y
=
dx x cos y − sin xsin 2 y16sin 2 y
(
)
5.
Find the equations of the lines tangent and normal to the curve
2
5x − 3xy + y2 = 3x at the point (1, 2) .
dy
dy
10x − 3x + y ( −3) + 2y = 3
dx
dx
dy
dy
10 (1) − 3(1) + ( 2 )( −3) + 2 ( 2 ) = 3
dx
dx
dy
= −1
dx
y − 2 = −1( x −1)
Tangent:
Normal:
y − 2 = 1( x −1)
2 2.3 Homework Solutions
1.
y = sin −1 ( ex )
1
ex
x
y′ =
⋅e =
1− e2 x
1− e2 x
3.
y = sin −1 ( 2x + 1)
y′ =
1
1− ( 2x +1)
2
⋅2 =
2
(
)
1− 4x 2 + 4x +1
y = xcos−1 x − 1− x 2
− 12
−1
1
−1
y′ = x ⋅
+
cos
x
−
1−
x
⋅ ( −2x )
(
)
2
1− x 2
−x
x
=
+
+ cos−1 x
2
2
1− x
1− x
−1
= cos x
5.
7.
(
y = sin−1 2 ( x )
1
y′ =
1−
9.
( )
2x
2
)
2
⋅ 2=
1−
dy
=0
dx
sec−1 x
y=
x
y′ =
2
⎛ 1⎞
y = cot −1 ⎜ ⎟ − tan−1 x
⎝ x⎠
y = tan −1 x − tan −1 x = 0 ⇒
11.
( )
2x
x⋅
1
1
− sec−1 x
− sec−1 x
2
2
x x −1
= x −1 2
x2
x
3 (
= 2 −4x − 4x
2
)
− 12
=
(−x
−1
2
−x
)
1
2
2.4 Homework Solutions
1.
Find the equation of the tangent line to f (x) = x 5 − 5x + 1 at x = −2 and use
it to get an approximate value of f (-1.9).
f ( −2 ) = −21
y + 21 = 75 ( x + 2 )
f ′ ( x ) = 5x 4 − 5 ⇒ f ′ ( −2 ) = 75
y = −21+ 75 ( x + 2 ) ⇒ y x=−1.9 = −13.5
f ( −1.9 ) ≈ −13.5
3.
Find all points on the graph of y = 2sin x + sin 2 x where the tangent line is
horizontal.
f ′ ( x ) = 2cos x + 2sin x cos x = 0
cos x = 0
or
⎧ π
x = ⎨ ± ± 2π n
⎩ 2
sin x = −1
⎧ 3π
x = ⎨ ± 2π n
⎩ 2
⎛π
⎞ ⎛ π
⎞
Points where the tangent line is horizontal: ⎜ ± 2π n, 3⎟ , ⎜ − ± 2π n, 1⎟
⎝2
⎠ ⎝ 2
⎠
5.
(
use it to approximate g (1.1) .
g (1) = 7
(
)
g′ ( x ) = 2 + sec2 x 2 −1 ( 2x ) ⇒ g′ (1) = 4
y − 7 = 4 ( x −1)
g (1.1) ≈ y (1.1) = 4 (1.1−1) + 7 = 7.4
)
Find the equation of the tangent line to g ( x ) = 5 + 2x + tan x 2 −1 at x = 1 and
4 7.
⎛ π 2⎞
x at x = 1 and
⎝ 2 ⎟⎠
Find the equation of the tangent line to f ( x ) = ln xsin ⎜
plug x = 1.1 into both the original and the tangent line equation. Explain why the
values are so similar.
f (1) = 0
⎛
⎛π ⎞ π⎞
⎛ π ⎞ ⎛ 1⎞
f ′ ( x ) = ( ln x ) ⎜ cos ⎜ x ⎟ ⎟ + sin ⎜ x ⎟ ⎜ ⎟ ⇒ f ′ (1) = 1
⎝ 2 ⎠ 2⎠
⎝ 2 ⎠ ⎝ x⎠
⎝
y = x −1
f (1.1) ≈ y (1.1) = .090
f (1.1) = .1
The results are so similar because of local linearity. The tangent line is extremely
close to the curve near the point of tangency, so values of y for the tangent line
serve as good approximations for values of f near the point of tangency.
5