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Math 11-Worksheet 07 Sol

Solutions

Course

Calculus I (MATH 011)

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University of California, Merced

Academic year: 2022/2023

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1 Section 3.

All I see are patterns: Take the derivative of each of the following functions. You might need the Chain Rule at first, but try to find an underlying pattern.

y = sin θ, y = sin 2θ, y = sin 3θ, y = sin 4θ, y = sin πθ

We find that if y = sin(kθ), then y′ = k cos(kθ).

All I see are patterns (again): Take the derivative of each of the following functions. You might need the Chain Rule at first, but try to find an underlying pattern.

y =

√

x, y =

√

2 + sin x, y =

√

3 + cos x, y =

√

4 + tan x, y =

√

5 + 6x 7

We find that if y =

√

f (x), then y ′ = 1 2 [

f (x)]

− 1 / 2 f ′ (x) or

f ′(x)

√

f (x).

Chain of fools: Take the derivative of the following functions. You may need to utilize the Chain Rule. It might be helpful to first decompose each expression first.

(a) A(x) = (4x − x 2 ) 100

A ′ (x) = 100(4 − 2 x)(4x − x 2 ) 99

(b) B(x) = (1 + x 4 ) 2 / 3

B′(x) = 8 3

x 3 (1 + x 4 ) − 1 / 3

√

1 + tan t

C′(t) = 1 3 sec

2 t(1 + tan t) − 2 / 3

(d) D(θ) = a 3 + cos 3 θ D′(θ) = −3 sin θ cos 2 θ

(e) E(θ) = 4 cos nθ E ′ (θ) = − 4 n sin(nθ)

(f) F (p) = (3p − 1) 4 (2p + 1) − 3

F ′(p) = 6(

p − 1) 3 (p + 3)

(2p + 1) 4

(g) G(s) =

√

s 2 + 1

s 2 + 4

G′(s) =

3 s

(s 2 + 1) 1 / 2 (s 2 + 4) 3 / 2

(h) H(x) =

x √ 7 − 3 x

H

′ (x) =

7 −

3 2 x (7 − 3 x) 3 / 2

(i) I(θ) = sin

(√

1 + 10θ

)

I

′ (θ) = 5 cos(

√

1 + 10θ) √ 1 + 10θ

(j) J(x) = cos 2 (sin x) J ′ (x) = −2 cos(sin x) sin(sin x) cos x

Decomposition: (i) Find functions f , g, and h such that F = f ◦ g ◦ h = f (g(h(x))). Note: there might be multiple answers (come up with at least one answer for each F ). (ii) Find the derivative, F ′ (x), of each.

(a) F (x) =

√

1 − √x f (x) =

√

x, g(x) = 1 − x, h(x) =

√

(b) F (x) = sin 3 (2x + 3) f (x) = x 3 , g(x) = sin x, h(x) = 2x + 3

1

(2x 2 + x + 3) 3 f (x) = 1/x, g(x) = x 3 , h(x) = 2x 2 + x + 3

(d) F (x) =

√

x + 1 − 1 √ x + 1 + 1

f (x) =

x − 1

x + 1 ,

g(x) =

√

h(x) = x + 1

Off on a tangent Find the equation of the tangent line to the curve at the given point.

(a) f (x) =

√

1 + x 3 , (2, 3) y = 2x − 1

(b) f (θ) = sin θ + sin

2 θ, (0, 0) y = x

(d) f (θ) = tan 2 θ, (π/ 4 , 1)

y = 4x + 1 − π

(e) f (x) = csc x, (π/ 6 , 2)

y = − 2

√

3 x + 2 +

√

3

(f) f (θ) = cot θ, (π/ 6 ,

√

2)

y = − 4 x +

√

2 + 2

2 Section 3.

Find the given quantity with the information provided.

(a)

dx f

− 1 (2), given f − 1 (2) = 0 and f ′ (0) = − 2

dx f

− 1 (2) = − 1 / 2

(b)

dx f

− 1 (6), given f ′(−1) = 6 and f − 1 (6) = − 1

dx f

− 1 (6) = 1/ 6

dx f

− 1 (0) = 2/3 and f − 1 (0) = 3 f ′ (3) = 3/ 2

Given the information and indicated function, find

dx f

− 1 (a) for the given value of a.

(a) f (x) = x 3 + x + 1 at a = 1. Note: f (0) = 1

Since f (0) = 1, then f − 1 (1) = 0. Using f ′(x) = 3x 2 + 1, we find that

dx f

− 1 (1) = 1.

(b) f (x) = 1 − x 2 − x 3 at a = 5. Note: f (−2) = 5.

Since f (−2) = 5, then f − 1 (5) = −2. Using f ′(x) = − 2 x − 3 x 2 , we find that

dx f

− 1 (5) = −

1

8.

x + 1

x − 2

at a = −3. Note: f (1) = −3..

Since f (1) = −3, then f − 1 (−3) = 1. Using f ′ (x) = −

5

(x − 2) 2

, we find that

dx f

− 1 (−3) = −5.

(d) f (x) =

− 3 x + 2

− 3 − x

at a = 14. Note: f (−4) = 14..

Since f (−4) = 14, then f − 1 (14) = −4. Using f ′(x) =

11

(− 3 − x) 2

, we find that

dx f

− 1 (14) = 11.

(e) f (x) =

√

− 1 − 5 x at a = 7. Note: f (−10) = 7..

Since f (−10) = 7, then f − 1 (7) = −10. Using f ′(x) = −

5 2 (

− 1 − 5 x) − 1 / 2 , we find that

dx f

− 1 (7) =

−

14

5.

(f) f (x) =

√

3 x + 4 at a = 2. Note: f (0) = 2..

Since f (0) = 2, then f − 1 (2) = 0. Using f ′ (x) = 3 2 (

x + 4) − 1 / 2 , we find that

dx f

− 1 (2) = 4 3.

Troubleshooting What is wrong with the following calculations? What should the derivative be?

y = 7

x 3

⇒ y

′ = 7

x (3x 2 ) − (ln 7) x x 3

x 6

The numerator is reversed, as the result should be

y ′ = (ln 7)

xx 3 − 7 x(3x 2 )

x 6

.

This can be further simplified to

y ′ = 7

(

(ln 7)x − 3

)

x 3

.

A probe leaving a planet has a position function given by r(t) = R 0 + et −

1

2

mgt 2 , where R 0 is the

radius of the planet, m is the mass of the probe, g is a constant related to the gravitational pull of the planet, and t is time, in seconds. Find the (a) velocity and (b) accleration functions for the probe. (a) v(t) = et + mgt and (b) a(t) = et + mg.

Compute the derivatives of the following functions.

(a) A(t) = 5t 2 + 4e t

A ′ (t) = 10t + 4e t

(b) B(x) = 12e x + 11 x

B ′ (x) = 12e x + (ln 11) 11 x

C ′ (x) = (ln 2) 2 x + (2 ln 3) 3 x

(d) D(x) = 3x − 2 · 4 x D′(x) = 3 − (2 ln 4) 4x

(e) E(x) = 3

3 + 33

√

E

′ (x) = ln 3 3

3

x −

33

2

x − 3 / 2

(f) F (x) = (ln 4) x

F ′(x) = (ln 4) 2 4 x

(g) G(t) = 5 · 5 t + 6 · 6 t

G ′ (t) = (5 ln 5) 5 t + (6 ln 6) 6 t

(h) H(x) = e 2 + x e

H ′ (x) = ex e− 1

(i) I(t) = π 2 + π t

I ′ (t) = (ln π) π t

(j) J(x) = π x + x π

J ′ (x) = (ln π) π x + πx π− 1

(k) K(x) = e 1+x

K ′ (x) = e 1+x

(l) L(θ) = e θ− 1

L ′ (θ) = e θ− 1

(m) M (x) = xπ

2 + (π 2 )x

M ′(x) = π 2 xπ

2 − 1 + (ln(π 2 )) (π 2 )x

Find the derivative of each of the following functions.

(a) A(t) = t 2018 e t

A ′ (t) = t 2017 e t (2018 + t)

(b) B(x) = x 2 25 x

B ′ (x) = x 25 x (2 + x ln 25)

y 2

C

′ (y) = 2

y (ln 2 − 2)

y 3

(d) D(x) =

e 2 x − 1

D

′ (x) = −

e x (e 2 x + 1)

(e 2 x − 1) 2

(e) E(t) = (

t + 1)e t

t 2

E′(t) =

et(t 2 − 2)

t 3 (f) F (y) = 2y · 3 y F ′(y) = (ln 2 + ln 3) 2y · 3 y

(g) G(x) = (x 3 − 2 x 2 − x + 1)(et − 2 t) (You do not have to simplify) G′(x) = (3x 2 − 4 x − 1)(et − 2 t) + (x 3 − 2 x 2 − x + 1)(et − (ln 2) 2t)

A missile launched from a naval vessel follows the flight path given by h(t) = 2 log(t 2 + 1), with h in

meters and t in seconds. Write the velocity function, v(t), for the missile. Use this to compute the missile’s velocity at t = 0 and t = 3 seconds.

v(t) = h′(t) =

4 t

(t 2 + 1) ln 10

⇒ v(0) = 0 and v(3) =

6

5 ln 10

Find the equation of the tangent line to the curve f (x) = 1 − ln(1 + 2x) at the point x = 0.

f ′ (x) = −

2

1 + 2x

⇒ f ′ (0) = −2. Using (0, 1), we get the tangent line, y = − 2 x + 1.

Compute the derivatives of the given functions.

(a) f (x) = 4 log 3 (2x 2 − x) − 3 log 4 (3 + x)

f ′ (x) =

4(4x − 1)

(2x 2 − x) ln 3

−

3

(3 + x) ln 4

(b) g(t) = log 5 t · log 5 (t + 1)

g′(t) =

1

t ln 5 log 5 (t + 1) +

1

(t + 1) ln 5 log 5 t

log(r + 1)

1 + log(1 − r)

h′(r) =

1

(r + 1) ln 10 (1 + log(

− r)) +

1

(1 − r) ln 10 log(

r + 1)

(1 + log(1 − r)) 2

(d) y(x) = ln(

√

1 − 2 x + x 2 )

y ′ (x) =

−1 + x √ ln(1 − 2 x + x 2 )

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Math 11-Worksheet 07 Sol

Course: Calculus I (MATH 011)

81 Documents

Students shared 81 documents in this course

University: University of California, Merced

Was this document helpful?

Math 11 Worksheet 7 for Sections 3.6, 3.7, and 3.9 (Solutions) Spring 2023

1 Section 3.6

1. All I see are patterns: Take the derivative of each of the following functions. You might need the

Chain Rule at first, but try to find an underlying pattern.

y= sin θ, y = sin 2θ, y = sin 3θ, y = sin 4θ, y = sin πθ

We find that if y= sin(kθ), then y′=kcos(kθ).

2. All I see are patterns (again): Take the derivative of each of the following functions. You might

need the Chain Rule at first, but try to find an underlying pattern.

y=√x, y =√2 + sin x, y =√3 + cos x, y =√4 + tan x, y =p5 + 6x7

We find that if y=pf(x), then y′=1

2[f(x)]−1/2f′(x) or f′(x)

2pf(x).

3. Chain of fools: Take the derivative of the following functions. You may need to utilize the Chain

Rule. It might be helpful to first decompose each expression first.

(a) A(x) = (4x−x2)100

A′(x) = 100(4 −2x)(4x−x2)99

(b) B(x) = (1 + x4)2/3

B′(x) = 8

3x3(1 + x4)−1/3

√1 + tan t

C′(t) = 1

3sec2t(1 + tan t)−2/3

(d) D(θ) = a3+ cos3θ

D′(θ) = −3 sin θcos2θ

(e) E(θ) = 4 cos nθ

E′(θ) = −4nsin(nθ)

(f) F(p) = (3p−1)4(2p+ 1)−3

F′(p) = 6(3p−1)3(p+ 3)

(2p+ 1)4

(g) G(s) = rs2+ 1

s2+ 4

G′(s) = 3s

(s2+ 1)1/2(s2+ 4)3/2

(h) H(x) = x

√7−3x

H′(x) = 7−3

(7 −3x)3/2

(i) I(θ) = sin √1 + 10θ

I′(θ) = 5 cos(√1 + 10θ)

√1 + 10θ

(j) J(x) = cos2(sin x)

J′(x) = −2 cos(sin x) sin(sin x) cos x

4. Decomposition: (i) Find functions f,g, and hsuch that F=f◦g◦h=f(g(h(x))). Note: there

might be multiple answers (come up with at least one answer for each F). (ii) Find the derivative,

F′(x), of each.

(a) F(x) = p1−√x

f(x) = √x,g(x) = 1 −x,

h(x) = √x

(b) F(x) = sin3(2x+ 3)

f(x) = x3,g(x) = sin x,

h(x) = 2x+ 3

(2x2+x+ 3)3

f(x) = 1/x,g(x) = x3,

h(x) = 2x2+x+ 3

(d) F(x) = √x+ 1 −1

√x+ 1 + 1

f(x) = x−1

x+ 1,g(x) = √x,

h(x) = x+ 1

Math 11-Worksheet 07 Sol

Calculus I (MATH 011)

University of California, Merced

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1 Section 3.

√

√

√

√

√

√

√

√

√

H

7 −

(√

)

I

√

√

√

√

1

√

√

√

√

√

3

3

√

2)

√

2 + 2

2 Section 3.

1

8.

5

11

√

5

2 (

−

14

5.

√

.

(

)

.

1

2

3 + 33

√

E

3

33

2

C

D

6

2

−

3

1

1

1

1

√

Math 11-Worksheet 07 Sol

Course: Calculus I (MATH 011)

University: University of California, Merced

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