HW 5 – Solutions
Calculus I
Part II – Instructor’s Problems
21. Solutions
3
4
a. 𝑦 = 5sec 3 (√𝑥 3 ) = 5[sec(𝑥 3/4 )]
2
3
𝑦 ′ = 3 ⋅ 5[sec(𝑥 3/4 )] ⋅ sec(𝑥 3/4 ) tan(𝑥 3/4 ) ⋅ 4 𝑥 −1/4
𝑦′ =
45 sec3(𝑥 3/4 ) tan(𝑥 3/4 )
4𝑥 1/4
𝑥 sin 𝑥
)
4𝑥
b. 𝑓(𝑥) = log 4 (
1
= log 4 𝑥 + log 4 sin 𝑥 + log 4 4𝑥 = log 4 𝑥 + log 4 sin 𝑥 + 𝑥
1
𝑓 ′ (𝑥) = 𝑥 ln 4 + (sin 𝑥) ln 4 ⋅ cos 𝑥 + 1
1
𝑓 ′ (𝑥) = 𝑥 ln 4 +
cot 𝑥
ln 4
+1
c. 𝑦 = arctan(ln 𝑥) − arcsin(𝑒 2𝑥 + 3)
1
1
𝑦 ′ = 1+(ln 𝑥)2 ⋅ 𝑥 −
1
𝑦 ′ = 𝑥+𝑥(ln 𝑥)2 −
1
√1−(𝑒 2𝑥 +3)2
⋅ 2𝑒 2𝑥
2𝑒 2𝑥
√1−(𝑒 2𝑥 +3)2
22. Solution
By the chain rule we know:
(𝑓 ∘ 𝑔)′ (𝑥) = 𝑓 ′ (𝑔(𝑥)) ⋅ 𝑔′ (𝑥)
so
(𝑓 ∘ 𝑔)′ (1) = 𝒇′ (𝒈(𝟏)) ⋅ 𝒈′ (𝟏)
From the table we know:
1. 𝒈′ (𝟏) = 𝟐
2. 𝒈(𝟏) = −𝟐
3. 𝒇′ (−𝟐) = 𝟏
Therefore:
(𝑓 ∘ 𝑔)′ (1) = 𝒇′ (−𝟐) ⋅ 𝟐 = 𝟏 ⋅ 𝟐 = 2
23. Solutions
5
𝑓(𝑥) =
5 √𝑥
(𝑥 2 +1)3
5𝑥 1/5
= (𝑥 2
+1)3
a. By quotient rule,
𝑓 ′ (𝑥) =
𝑥 −4/5 ⋅(𝑥 2 +1)3 −3(𝑥 2 +1)2 ⋅2𝑥⋅5𝑥 1/5
((𝑥 2 +1)3 )2
=
𝑥 −4/5 ⋅(𝑥 2 +1)3 −30𝑥 6/5 (𝑥 2 +1)2
(𝑥 2 +1)6
.
Then we factor.
𝑓 ′ (𝑥) =
𝑥 −4/5(𝑥 2 +1)2 [(𝑥 2 +1)1 −30𝑥 10/5 ]
(𝑥 2 +1)6
=
1−29 𝑥 2
𝑥 4/5 (𝑥 2 +1)4
.
b. To find where the function has a horizontal tangent line, set the numerator of the derivative
to zero.
1
1
1 − 29𝑥 2 = 0 → 𝑥 2 = 29 → 𝑥 = ±√29
c. To find where the function has a vertical tangent line, set the denominator of the derivative
to zero.
𝑥 4/5 = 0 or (𝑥 2 + 1)4 = 0
𝑥 = 0 or 𝑥 = √−1 (not real)
24. Solutions
a. ℎ(0) = 3 − cos(2 ⋅ 0) = 2 feet
b. 𝑣(𝑡) = ℎ′ (𝑡) = 2 sin 2𝑡
1
c. 𝐾(𝑡) = 2 𝑚(2 sin 2𝑡)2 = 2𝑚 sin2 2𝑡
And then by the chain rule
formula for sine).
𝑑𝐾
𝑑𝑡
= 8𝑚 sin 2𝑡 ⋅ cos 2𝑡 = 4𝑚 sin 4𝑡 (if you use the double angle
25. Solution
𝑑
(𝑦 2
𝑑𝑥
𝑑
− 5𝑥𝑦 + 𝑥 2 ) = 𝑑𝑥 (1)
𝑑𝑦
𝑑𝑦
2𝑦 ⋅ 𝑑𝑥 − 5 (1 ⋅ 𝑦 + 𝑑𝑥 ⋅ 𝑥) + 2𝑥 = 0
𝑑𝑦
𝑑𝑦
2𝑦 𝑑𝑥 − 5𝑦 − 5𝑥 𝑑𝑥 + 2𝑥 = 0
𝑑𝑦
𝑑𝑦
2𝑦 𝑑𝑥 − 5𝑥 𝑑𝑥 = 5𝑦 − 2𝑥
𝑑𝑦
(2𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
− 5𝑥) = 5𝑦 − 2𝑥
5𝑦−2𝑥
= 2𝑦−5𝑥
𝑑𝑦
𝑚 = 𝑑𝑥 |
5(5)−2(1)
(1,5)
= 2(5)−5(1) =
23
5
Next we use point-slope form and plug in the given point as well as the slope we just found.
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
𝑦−5=
𝑦=
23
𝑥
5
23
(𝑥
5
2
−5
− 1)