Test 2 Review Solutions
Math 1910
1. Differentiate and simplify:
a. y = x 4 cos x
y ′ = x 4 (− sin x ) + (cos x)(4x 3 ) = x 3 (4 cosx − x sin x )
b. y =
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3x − 4
2x + 1
y=
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c. y = sec(5x 2 + 3x + 2)
2
d. y = cot x = (cot x )
2
(2x + 1)( 3) − (3x − 4)(2) 6x + 3− 6x + 8
11
=
=
2
2
2
(2x + 1)
(2x + 1)
(2x + 1)
y ′ = (10x + 3) sec(5x 2 + 3x + 2)tan(5x 2 + 3x + 2)
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y ′ = 2(cot x )(− csc2 x ) = −2 cot x csc 2 x
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2
2
e. y = x sin( x + 1)
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[
€
€ ( x 2 + 1)(2x) = 2x x 2 cos( x 2 + 1) + sin( x 2 + 1)
y ′ = x 2 cos( x 2 + 1)(2x ) + sin
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f. f (x) = x sin x tan x
]
f ′(x) = sin x tan x + x cos x tan x + x sin x sec 2 x
dy
implicitly:
dx
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4
6
a. x sin y + y = x
x cos yy ′ + sin y + 4y3 y′ = 6x 5
2. Find
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b. y − tan y − sin x = 0
y ′ − sec 2 y y′ − cos x = 0
x cos yy ′ + 4y3 y′ = 6x 5 − sin y
( x cos y + 4y ) y′ = 6x
3
y′ =
3. Find
5
y ′ − sec 2 y y ′ = cos x
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− sin y
y′(1− sec 2 y ) = cos x
6x 5 − sin y
x cos y + 4y3
y′ =
cos x
cos x
=−
2
1 − sec y
tan 2 y
d 2y
2
2 : x+ y = y
dx
a. y = sin(2x + 1)
y€′ = cos(2x + 1)(2) = 2cos(2x + 1)
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y ′′ = −2sin(2x + 1)(2) = −4 sin(2x + 1)
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2
b. x + y = y (Use implicit differentiation.)
1
−1
1 + y′ = 2yy ′
y′ =
= (2y − 1)
2y − 1
1 = 2yy ′ − y ′
1 = y′( 2y − 1)
y′ =
1
2y − 1
y ′′ = −1( 2y − 1)
−2
(2 y′ )
 2 
−2
2
 = −
= −1( 2y − 1) 
(2y − 1)3
 2y − 1 
4. Let P(x) = f (x)g(x), where the
functions f and g are graphed in
the diagram on the right.
y
y = f (x)
Find P′(1).
P′(1) = f ′(1)g(1) + f (1)g ′(1)
 1
 3
=  −  (−3) +   (−1) = 0
 2
 2
x
y = g(x)
5. Find all values of x for which the functions have horizontal tangents:
7
4x − 3
3
a. f (x) = x (3x − 2)
b. g(x) = 2
2x + 1
2
6
7
2x
+ 1)( 4) − (4x − 3)(4x )
€ (3x − 2) (3) + 3x 2 (3x − 2) g ′(x) = (
f ′(x) = x 3 (7)
6
7
( 2x 2 + 1) 2
= 21x 3 ( 3x − 2) + 3x 2 ( 3x − 2)
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6
−8x 2 + 12x + 4
2
= 3x ( 3x − 2) [ 7x + (3x − 2)]
=
2
(2x 2 + 1)
6
2
= 3x ( 3x − 2) (10x − 2)
g ′(x) = 0 ⇒ −8x 2 + 12x + 4 = 0
2
1
f ′(x) = 0 ⇒ x = 0, x = ,x =
3
5
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6. Find all critical numbers:
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3
a. f (x) = x − x + 2
f ′(x) = 3x 2 − 1
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2
3x − 1 = 0
2
3x = 1
1
x2 =
3
x=±
1
3
1
+ 81x = x −1 + 81x
x
g′ (x) = −1x −2 + 81
b. g(x) =
−1x −2 + 81 = 0
1
= 81
x2
1
x2 =
81
1
x=±
9
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−8x 2 + 12x + 4 = 0
−4(2x 2 − 3x − 1) = 0
2x 2 − 3x − 1 = 0
3 ± (−3)2 − 4(2)(−1)
x=
2(2)
x=
3 ± 17
4
c. y = x + sinx
y′ = 1 + cos x
1 + cos x = 0
cos x = −1
x = (2k + 1)π
for all integers k
7. Consider the function f (x) = x 3 − 3x + 7 on the interval [0, 4].
a. Find the absolute extrema on the the given interval.
f ′(x) = 3x 2 − 3
f (0) = 7
3x 2 − 3 = 0 €
x = ±1
f (1) = 5
f (4) = 59
Absolute maximum at (4,49) and absolute minimum at (1,5)
b. Find the value of c guaranteed by the Mean Value Theorem.
3c 2 − 3 = 13
f (4) − f (0) 59 − 7
=
= 13
16
4
4−0
4
c=
=
3
3
8. Does the function f (x) = x 2 − 5x + 3 on the interval [1,3] satisfy the criteria of Rolle's theorem?
No: f (1) ≠ f (3).
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9. Use the quotient rule to derive the formula for the derivatives of the functions y = csc x and t = tan x .
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y = csc x
1
=
sin x
(sin x )(0) − (1)(cos x )
y′ =
sin 2 x
1 cosx
=−
sin x sin x
= − csc x cot x
10. Find the second derivatives:
a. f (x) = sin x + 2x 3
f ′(x) = cos x + 6x 2
f ′′(x) = − sin x + 12x
y = tan x
sin x
=
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cos x
(cos x )(cosx) − (sin x )( −sin x )
y′ =
cos2 x
cos 2 x + sin 2 x
=
cos2 x
1
=
= sec 2 x
2
cos x
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1/ 2
b. g(x) = x 2 + 1 = ( x 2 + 1)
−1/ 2
−1/ 2
1
g ′(x) = ( x 2 + 1) (2x ) = x ( x 2 + 1)
2
 1
−3/ 2
−1 / 2
g ′′(x) = x  −  ( x 2 + 1) (2x) + (x 2 + 1)
 2
g ′′(x) = −x ( x + 1)
2
g ′′(x) = −
2
−3 / 2
x2
(x
2
+ 1)
3 /2
+
+ ( x + 1)
2
x2 + 1
(x
2
+ 1)
3 /2
−1/ 2
=
=−
x2
(x
2
+ 1)
1
(x
2
+ 1)
3/ 2
3/ 2
+
1
(x
2
+ 1)
1/ 2