Math 151 Test 2
Students Name:___________________
1.
Draw on the grid below a sketch of f (x) by using the given graph of f(x).
y = f(x)
y=
2.
Differentiate the following functions.
(a) f(x)
=
(b) f(x)
=
(c) g(x)
=
5𝑥 3 − 4√𝑥 +
6
√𝑥
(𝑥 3 − 6𝑥 2 ) 𝑒 𝑥
𝑥 2 +1
𝑥 2 −1
2
+ 𝑥3 − 𝜋 3
3.
Stone is thrown vertically upwards from the surface of a planet, its height in meters after t
seconds is given by the formula h = t4 – 4t3
(a) What is the velocity and the acceleration of the stone after 3 seconds?
(b) At what time is the stone stationary?
(c) What is the velocity of the stone when it hits the ground?
(d) Is the function speeding up or slowing down at t = 4 seconds.
4.
Differentiate the following functions and simplify as much as possible.
(a) h(x) =
(b) g(x)
𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥
𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥
=
sin x + 2cos x – 3sec x + 4tanx + 5csc x – 6cot x
5.
Differentiate the following functions using the chain rule and simplify as much as possible.
(a)
p(x)
(b)
y
=
(c)
y
=
sin(𝑒 𝑥 + 𝑒 −𝑥 )
=
sin5 ( cos(4x + 1) )
4𝑥
√𝑥 2 +1
6.(a)
Use implicit differentiation to differentiate x2y3 – sin y = x4 – 1
6.(b)
Find the equation of the tangent to x2y3 – sin y = x4 – 1 at the point (1,0)
𝑑𝑦
𝑑2 𝑦
7.
Find 𝑑𝑥 and 𝑑𝑥 2 for x3 + y3 = 4 and simplify as much as possible.
8.
Bonus Question (5 extra points) Differentiate f(x) = 𝑎sin(bcos(ctan(𝑒 𝑑𝑥 )))
Solutions
1.
Draw on the grid below a sketch of f (x) by using the given graph of f(x).
y = f(x)
y=
2.
Differentiate the following functions.
6
2
(a) f(x)
=
5𝑥 3 − 4√𝑥 +
f(x)
=
5𝑥 3 − 4𝑥 2 + 6𝑥 −2 + 2𝑥 −3 − 𝜋 3
𝑓′(𝑥)
=
15𝑥 2 − 2𝑥 −2 − 3𝑥 −2 − 6𝑥 −4
𝑓′(𝑥)
=
15𝑥 2 −
√𝑥
+ 𝑥3 − 𝜋 3
1
1
1
2
√𝑥
3
3
6
− √𝑥 3 − 𝑥 4
=
(𝑥 3 − 6𝑥 2 ) 𝑒 𝑥
𝑓′(𝑥)
=
𝑓 ′ (𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔′(𝑥)
𝑓′(𝑥)
=
(3𝑥 2 − 12𝑥)𝑒 𝑥 + (𝑥 3 − 6𝑥 2 ) 𝑒 𝑥
𝑓′(𝑥)
=
3𝑥 2 𝑒 𝑥 − 12𝑥𝑒 𝑥 + 𝑥 3 𝑒 𝑥 − 6𝑥 2 𝑒 𝑥
𝑓′(𝑥)
=
𝑥 3 𝑒 𝑥 − 3𝑥 2 𝑒 𝑥 − 12𝑥𝑒 𝑥
𝑓′(𝑥)
=
𝑥𝑒 𝑥 (𝑥 2 − 3𝑥 − 12)
(b) f(x)
(c) g(x)
=
𝑥 2 +1
𝑥 2 −1
𝑓 ′ (𝑥)𝑔(𝑥)−𝑓(𝑥)𝑔′(𝑥)
(𝑔(𝑥))2
𝑔′(𝑥)
=
𝑔′(𝑥)
=
𝑔′(𝑥)
=
2𝑥 3 −2𝑥−2𝑥 3 −2𝑥
(𝑥 2 +1)2
𝑔′(𝑥)
=
−4𝑥
(𝑥 2 +1)2
2𝑥(𝑥 2 −1)−(𝑥 2 +1)2𝑥
(𝑥 2 +1)2
3.
Stone is thrown vertically upwards from the surface of a planet, its height in meters after t
seconds is given by the formula h = t4 – 4t3
(a) What is the velocity and the acceleration of the stone after 3 seconds?
=
t4 – 4t3
=
4t3 – 12t2
V(t) =
4t3 – 12t2
V(3)
=
a(t) =
𝑉′(𝑡)
a(t) =
12t2 – 24t
a(3) =
12(3)2 – 24(3)
h
𝑑ℎ
𝑑𝑡
4(3)3 – 12(3)2
=
=
0 m/sec
36 m/sec2
(b) At what time is the stone stationary?
Stationary when
V(t) =
0
V(t) =
4t3 – 12t2
4t2 (t– 3)
=
=
0
0
Stationary at t = 0 seconds and t = 3 seconds
(c) What is the velocity of the stone when it hits the ground?
Hits the ground when
h
4
3
t – 4t
t3(t– 4)
=
0
=
=
0
0
It will hit the ground when t = 4 seconds
V(4)
=
4(4)3 – 12(4)2 =
64 m/sec
(d) Is the function speeding up or slowing down at t = 4 seconds.
V(4)
=
4(4)3 – 12(4)2 =
64 m/sec
a(4)
=
12(4)2 – 24(4) =
96 m/sec2
Object is speeding up at t = 4 seconds
4.
Differentiate the following functions and simplify as much as possible.
𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥
(a) h(x)
=
h(x)
=
ℎ′(𝑥)
=
𝑓 ′ (𝑥)𝑔(𝑥)−𝑓(𝑥)𝑔′(𝑥)
(𝑔(𝑥))2
ℎ′(𝑥)
=
(𝑐𝑜𝑠𝑥−𝑠𝑖𝑛𝑥)(sin 𝑥−𝑐𝑜𝑠𝑥)−(𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥)(𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥)
(𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥)2
ℎ′(𝑥)
=
ℎ′(𝑥)
=
ℎ′(𝑥)
=
ℎ′(𝑥)
=
ℎ′(𝑥)
=
(b) g(x)
𝑔′(𝑥)
𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥
𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥
𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥
𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥−𝑐𝑜𝑠2 𝑥−𝑠𝑖𝑛2 𝑥+𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑥−𝑠𝑖𝑛2 𝑥−𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥−𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥−𝑐𝑜𝑠2 𝑥
(𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥)(𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥)
−2𝑠𝑖𝑛2 𝑥−2𝑐𝑜𝑠2 𝑥
(𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥)(𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥)
−2
𝑠𝑖𝑛2 𝑥−𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥−𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥+𝑐𝑜𝑠2 𝑥
−2
1−2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥
−2
1−𝑠𝑖𝑛2𝑥
=
sin x + 2cos x – 3sec x + 4tanx + 5csc x – 6cot x
=
cos x – 2sin x – 3sec x tanx + 4 sec2x – 5csc x cotx + 6 csc2x
5.
Differentiate the following functions and simplify as much as possible.
(a)
p(x)
𝑝′(𝑥)
𝑝′(𝑥)
=
=
=
sin(𝑒 𝑥 + 𝑒 −𝑥 )
cos(𝑒 𝑥 + 𝑒 −𝑥 ) ∙ (𝑒 𝑥 − 𝑒 −𝑥 )
(𝑒 𝑥 − 𝑒 −𝑥 )cos(𝑒 𝑥 + 𝑒 −𝑥 )
(b)
y
=
=
sin5 ( cos(4x + 1) )
5 sin4 ( cos(4x + 1)) –sin(4x + 1) 4
=
–20 sin(4x + 1)sin4 ( cos(4x + 1))
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
(c)
y
=
y
=
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
4𝑥
√𝑥 2 +1
4𝑥
1
(𝑥 2 +1)2
𝑓 ′ (𝑥)𝑔(𝑥)−𝑓(𝑥)𝑔′(𝑥)
(𝑔(𝑥))2
1
=
=
2
(𝑥 2 +1)
1
1
−
4(𝑥 2 +1)2 −4𝑥 2 (𝑥2 +1) 2
(𝑥 2 +1)
1
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
=
=
1
−
1
4(𝑥 2 +1)2 −4𝑥 (𝑥 2 +1) 2 2𝑥
−
4(𝑥 2 +1) 2 (𝑥 2 +1−𝑥 2 )
(𝑥 2 +1)
4(1)
3
(𝑥 2 +1)2
4
√(𝑥 2 +1)3
6.(a)
Use implicit differentiation to differentiate x2y3 – sin y = x4 – 1
x2y3 – sin y =
x4 – 1
x2y3 – sin y =
x4 – 1
𝑑𝑦
𝑑𝑦
𝑑𝑦
𝑑𝑦
𝑓 ′ (𝑥)𝑔(𝑦) + 𝑓(𝑥)𝑔′(𝑦) 𝑑𝑥 – cosy𝑑𝑥
2𝑥𝑦 3 + 𝑥 2 3𝑦 2 𝑑𝑥 – cosy𝑑𝑥
𝑑𝑦
𝑑𝑦
𝑥 2 3𝑦 2 𝑑𝑥 – cosy𝑑𝑥
𝑑𝑦
𝑑𝑥
(3𝑥 2 𝑦 2 – cosy )
𝑑𝑦
𝑑𝑥
6.(b)
=
4x3
=
4x3
=
4x3 – 2xy3
=
4x3 – 2xy3
=
4𝑥 3 −2𝑥𝑦 3
3𝑥 2 𝑦 2 −𝑐𝑜𝑠𝑦
2𝑥(2𝑥 2 −𝑦 3 )
= 3𝑥 2 𝑦 2 −𝑐𝑜𝑠𝑦
Find the equation of the tangent to x2y3 – sin y = x4 – 1 at the point (1,0)
Slope of tangent line at the point (1,0)
𝑑𝑦
𝑑𝑥
=
=
=
𝑑𝑦
𝑑𝑥
=
2𝑥(2𝑥 2 −𝑦 3 )
3𝑥 2 𝑦 2 −𝑐𝑜𝑠𝑦
2(1)(2(12 )−03 )
0−cos(0)
4−0
0−1
–4
Equation of the Tangent line at the point (1,0) with a slope of – 2
𝑦 − 𝑦1 =
𝑚(𝑥 − 𝑥1 )
y–0
=
−4(𝑥 − 1)
=
– 4x + 4
y
7.
𝑑2 𝑦
𝑑𝑦
Find 𝑑𝑥 and 𝑑𝑥 2 for x3 + y3 = 4 and simplify as much as possible.
x3 + y3 =
𝑑𝑦
3x2 + 3y2𝑑𝑥 =
4
0
3y2𝑑𝑥
– 3x2
𝑑𝑦
=
𝑑𝑦
=
𝑑𝑥
𝑑𝑦
=
𝑑𝑥
𝑑2 𝑦
=
𝑑𝑥 2
=
=
−3𝑥 2
3𝑦 2
−𝑥 2
𝑦2
𝑓 ′ 𝑔−𝑓𝑔′
𝑔2
−2𝑥𝑦 2 −(−𝑥 2 )2𝑦
(𝑦 2 )2
−2𝑥𝑦 2 +2𝑦𝑥 2
𝑦4
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
−𝑥2
=
−2𝑥𝑦 2 +2𝑦𝑥 2 2
𝑦
𝑦4
−2𝑥𝑦 2 −
=
𝑦4
−2𝑥𝑦 2 −
=
=
=
=
𝑑2 𝑦
𝑑𝑥 2
8.
=
2𝑥4
𝑦
2𝑥4
𝑦
𝑦4
𝑦
𝑦
−2𝑥𝑦 3 −2𝑥 4
𝑦5
−2𝑥(𝑦 3 +𝑥 3 )
𝑦5
−2𝑥(4)
𝑦5
−8𝑥
𝑦5
Bonus Question (5 extra points) Differentiate f(x) = 𝑎sin(bcos(ctan(𝑒 𝑑𝑥 )))
𝑓′(𝑥)
𝑓′(𝑥)
=
=
𝑎 cos(bcos(𝑐 tan(𝑒 𝑑𝑥 ))) ∙ −𝑏sin(ctan(𝑒 𝑑𝑥 )) ∙ 𝑐𝑠𝑒𝑐 2 (𝑒 𝑑𝑥 ) ∙ 𝑒 𝑑𝑥 ∙ 𝑑
−𝑎𝑏𝑐𝑑 𝑒 𝑑𝑥 𝑠𝑒𝑐 2 (𝑒 𝑑𝑥 )sin(𝑐 tan(𝑒 𝑑𝑥 ))cos(𝑏 cos(ctan(𝑒 𝑑𝑥 )))