MATH 12002 - CALCULUS I 2.3, 2.4, and 2.5: Computing Derivatives - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §2.3, §2.4, and §2.5: Computing Derivatives (Part 2)

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 12

Derivatives

10 f (x) = sin(x2 + 2)

Use the Chain Rule: f ′(x) = [cos(x2 + 2)] · d

dx (x2 + 2)

• =

[cos(x2 + 2)] · 2x = 2x cos(x2 + 2).

D.L. White (Kent State University) 2 / 12

Derivatives

11 f (x) = tan(2x − x3)

Use the Chain Rule: f ′(x) = [sec2(2x − x3)] · d

dx (2x − x3)

• =

[sec2(2x − x3)] · (2 − 3x2) = (2 − 3x2) sec2(2x − x3).

D.L. White (Kent State University) 3 / 12

Derivatives

12 f (x) = x cos x

Use the Product Rule: f ′(x) = d

dx x

• cos x + x ·

d

dx cos x

• =

1 · cos x + x(− sin x) = cos x − x sin x.

D.L. White (Kent State University) 4 / 12

Derivatives

13 f (x) = 4 sec7(2 − 4x)

First rewrite the function as f (x) = 4[sec(2 − 4x)]7, and then use the Chain Rule twice: f ′(x) = 4 · 7[sec(2 − 4x)]6 · d

dx sec(2 − 4x)

• =

4 · 7[sec(2 − 4x)]6 · [sec(2 − 4x) tan(2 − 4x)] · d

dx (2 − 4x)

• =

4 · 7[sec(2 − 4x)]6 · [sec(2 − 4x) tan(2 − 4x)] · (−4) = 28[sec6(2 − 4x)] · [sec(2 − 4x) tan(2 − 4x)] · (−4).

D.L. White (Kent State University) 5 / 12

Derivatives

14 f (x) = (tan x)(sec x)

Use the Product Rule: f ′(x) = d

dx tan x

• (sec x) + (tan x)

d

dx sec x

• =

(sec2 x)(sec x) + (tan x)(sec x tan x).

D.L. White (Kent State University) 6 / 12

Derivatives

15 f (x) = cos2 x − sin2 x

It may help to rewrite the function as f (x) = (cos x)2 − (sin x)2. Use the Chain Rule on each term: f ′(x) = 2(cos x)1 d

dx cos x

• − 2(sin x)1 d

dx sin x

• =

2(cos x)(− sin x) − 2(sin x)(cos x) = −4(sin x)(cos x). Note: f (x) = cos2 x − sin2 x = 1 − 2 sin2 x = 2 cos2 x − 1. Try computing the derivative using these other two forms of f (x).

D.L. White (Kent State University) 7 / 12

Derivatives

16 f (x) = 1 + tan x

1 − tan x Use the Quotient Rule: f ′(x) = d

dx (1 + tan x)

• (1 − tan x) − (1 + tan x)

d

dx (1 − tan x)

• (1 − tan x)2

= (sec2 x)(1 − tan x) − (1 + tan x)(− sec2 x) (1 − tan x)2 .

D.L. White (Kent State University) 8 / 12

Derivatives

17 f (x) = sec x

x5 Use the Quotient Rule: f ′(x) = d

dx sec x

• · x5 − (sec x) ·

d

dx x5

(x5)2 . = (sec x tan x) · x5 − (sec x) · 5x4 (x5)2 = (sec x tan x) · x5 − (sec x) · 5x4 x10 .

D.L. White (Kent State University) 9 / 12

Derivatives

18 f (x) = sin(x3 + 2)

cos(x3 + 2) Use the Quotient Rule, with the Chain Rule for the numerator and denominator: f ′(x) =

d dx sin(x3+2)

• cos(x3+2)−[sin(x3+2)]

d dx cos(x3+2)

• [cos(x3+2)]2

=

[cos(x3+2)]· d dx (x3+2)

• cos(x3+2)−[sin(x3+2)][− sin(x3+2)]·

d dx (x3+2)

• [cos(x3+2)]2

=

[cos(x3+2)]·3x2 cos(x3+2)−[sin(x3+2)][− sin(x3+2)]·3x2 [cos(x3+2)]2

. Note: f (x) = sin(x3+2)

cos(x3+2) = tan(x3 + 2).

Try computing the derivative using this other form of f (x).

D.L. White (Kent State University) 10 / 12

Derivatives

19 f (x) = sin √x +

√ sin x Observe that sin √x = sin(x1/2) and √ sin x = (sin x)1/2. Then use the Chain Rule on each term: f ′(x) = (cos √x) · d

dx

√x

• + 1

2(sin x)−1/2 ·

d

dx sin x

• =

(cos √x) · 1

2x−1/2 + 1 2(sin x)−1/2 cos x.

D.L. White (Kent State University) 11 / 12

Derivatives

20 f (x) = sin(tan x + sec x)

Use the Chain Rule: f ′(x) = [cos(tan x + sec x)] · d

dx (tan x + sec x)

• =

[cos(tan x + sec x)](sec2 x + sec x tan x).

D.L. White (Kent State University) 12 / 12