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Trigonometric Differentiation — Exercise Set I., Slides of Trigonometry

Stanford UniversityTrigonometry

Trigonometric Differentiation Exercises-1. Universitas ... Find the derivative f (x) for the following functions. ... f (x) = x3 tan(2x) − x2 sec(3x).

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Trigonometric Differentiation Exercises-1
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2003 Doug MacLean
Trigonometric Differentiation — Exercise Set I.
Find the derivative f(x) for the following functions. Say what you can about the sign of f(x).
(I.1) f(x) =tan3(4x) Solution
(I.2) f(x) =tan (4x)3Solution
(I.3) f(x) =cot √xSolution
(I.4) f(x) =sec2(2x+1)2Solution
(I.5) f(x) =csc3x−sec3xSolution
(I.6) f(x) =x3tan(2x) −x2sec(3x) Solution
(I.7) f(x) =sec2xcsc xSolution
(I.8) f(x) =xtan 1
xSolution
(I.9) f(x) =x2tan 1
xSolution
(I.10) f(x) =x3tan 1
xSolution
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Partial preview of the text

Download Trigonometric Differentiation — Exercise Set I. and more Slides Trigonometry in PDF only on Docsity!

Universitas

Sask atchewanensis

DEOETPAT-RIÆ

2003 Doug MacLean

Trigonometric Differentiation — Exercise Set I.

Find the derivative f ′ (x) for the following functions. Say what you can about the sign of f ′ (x).

(I.1) f (x) = tan^3 ( 4 x) Solution

(I.2) f (x) = tan

( 4 x)^3

Solution

(I.3) f (x) = cot

x

Solution

(I.4) f (x) = sec^2

( 2 x + 1 )^2

Solution

(I.5) f (x) = csc^3 x − sec^3 x Solution

(I.6) f (x) = x^3 tan ( 2 x) − x^2 sec ( 3 x) Solution

(I.7) f (x) = sec^2 x csc x Solution

(I.8) f (x) = x tan

x

Solution

(I.9) f (x) = x^2 tan

x

Solution

(I.10) f (x) = x^3 tan

x

Solution

Universitas

Sask atchewanensis

DEOETPAT-RIÆ

2003 Doug MacLean

Trigonometric Differentiation — Exercise Set II.

Find the derivative f ′ (x) for the following functions. Say what you can about the sign of f ′ (x).

(II.1) f (x) = x tan x Solution

(II.2) f (x) = tan^2 x Solution

(II.3) f (x) = tan^3 x Solution

(II.4) f (x) = tan^4 x Solution

(II.5) f (x) = tan^5 x Solution

(II.6) f (x) =

1 − cot x 1 + cot x Solution

(II.7) f (x) =

1 + cot x 1 − cot x Solution

(II.8) f (x) =

1 − tan x 1 + tan x Solution

(II.9) f (x) =

1 + tan x 1 − tan x

Solution

(II.10) f (x) =

tan x 1 − tan x Solution

Universitas

Sask atchewanensis

DEOETPAT-RIÆ

2003 Doug MacLean

Solution Set I

Back to Questions

(I.1) f (x) = tan^3 ( 4 x)

Solution: f ′ (x) = 3 tan^2 ( 4 x)( tan 4 x) ′^ = 3 tan^2 ( 4 x) sec 2 ( 4 x)( 4 x) ′^ = 12 tan^2 ( 4 x) sec 2 ( 4 x) ≥ 0

(I.2) f (x) = tan

( 4 x)^3

Solution: f ′ (x) = sec^2

( 4 x)^3

( 4 x)^3

= sec^2

( 4 x)^3

3 ( 4 x)^2

( 4 x) ′^ = sec^2

( 4 x)^3

3 ( 4 x)^2

12 ( 4 x)^2 sec^2

( 4 x)^3

(I.3) f (x) = cot

x

Solution: f (x) = cot

x

f ′ (x) = − csc 2

x

1 2

x

1 2

= − csc^2

x

1 2

x −^

1 (^2) = − 1 2

x csc^2

x

Universitas

Sask atchewanensis

DEOETPAT-RIÆ

2003 Doug MacLean

Back to Questions

(I.4) f (x) = sec^2

( 2 x + 1 )^2

Solution: f ′ (x) = 2 sec

( 2 x + 1 )^2

sec

( 2 x + 1 )^2

2 sec

( 2 x + 1 )^2

sec

( 2 x + 1 )^2

tan

( 2 x + 1 )^2

( 2 x + 1 )^2

2 sec^2

( 2 x + 1 )^2

tan

( 2 x + 1 )^2

2 ( 2 x + 1 )( 2 x + 1 ) ′^ =

2 sec 2

( 2 x + 1 )^2

tan

( 2 x + 1 )^2

2 ( 2 x + 1 )( 2 ) = 8 ( 2 x + 1 ) sec^2

( 2 x + 1 )^2

tan

( 2 x + 1 )^2

(I.5) f (x) = csc^3 x − sec^3 x

Solution: f ′ (x) = 3 csc^2 x( csc x) ′^ − 3 sec^2 x( sec x) ′^ =

3 csc^2 x( − csc x cot x) − 3 sec^2 x( sec x tan x) = −3 csc^3 x cot x − 3 sec^3 x tan x

(I.6) f (x) = x^3 tan ( 2 x) − x^2 sec ( 3 x)

Solution: f ′ (x) = (x^3 ) ′^ tan ( 2 x) + x^3 ( tan ( 2 x)) ′^ − (x^2 ) ′^ sec ( 3 x) − x^2 ( sec ( 3 x)) ′^ =

( 3 x^2 ) tan ( 2 x) + x^3 ( sec^2 ( 2 x))( 2 x) ′^ − ( 2 x) sec ( 3 x) − x^2 ( sec ( 3 x) tan ( 3 x))( 3 x) ′^ = ( 3 x^2 ) tan ( 2 x) + x^3 ( sec^2 ( 2 x))( 2 ) − ( 2 x) sec ( 3 x) − x^2 ( sec ( 3 x) tan ( 3 x))( 3 ) = ( 3 x^2 ) tan ( 2 x) + 2 x^3 ( sec^2 ( 2 x)) − 2 x sec ( 3 x) − 3 x^2 ( sec ( 3 x) tan ( 3 x))

Universitas

Sask atchewanensis

DEOETPAT-RIÆ

2003 Doug MacLean

Back to Questions

(I.10) f (x) = x^3 tan

x

Solution: f ′ (x) = (x^3 ) ′^ tan

x

  • x^3

tan

x

= 3 x^2 tan

x

  • x^3 sec^2

x

x

3 x^2 tan

x

  • x^3 sec^2

x

x^2 = 3 x^2 tan

x

− x sec^2

x

Universitas

Sask atchewanensis

DEOETPAT-RIÆ

2003 Doug MacLean

Solution Set II

Back to Questions

(II.1) f (x) = x tan x

Solution: f ′ (x) = (x) ′^ tan x + x( tan x) ′^ = tan x + x sec^2 x

(II.2) f (x) = tan^2 x

Solution: f ′ (x) = 2 tan x( tan x) ′^ = 2 tan x sec 2 x

(II.3) f (x) = tan^3 x

Solution: Back to Questions

f ′ (x) = 3 tan 2 x( tan x) ′^ = 3 tan^2 x sec 2 x > 0

(II.4) f (x) = tan^4 x

Solution: f ′ (x) = 4 tan^3 x( tan x) ′^ = 4 tan^3 x sec 2 x

Universitas

Sask atchewanensis

DEOETPAT-RIÆ

(II.8) f (x) =  2003 Doug MacLean

1 − tan x 1 + tan x

Solution: Back to Questions

f ′ (x) = ( 1 − tan x) ′ ( 1 + tan x) − ( 1 − tan x)( 1 + tan x) ′ ( 1 + tan x)^2

− sec 2 x( 1 + tan x) − ( 1 − tan x)( sec^2 x) ( 1 + tan x)^2

sec 2 x − ( 1 + tan x) − ( 1 − tan x) ( 1 + tan x)^2

−2 sec^2 x ( 1 + tan x)^2

(II.9) f (x) =

1 + tan x 1 − tan x

f ′ (x) = ( 1 + tan x) ′ ( 1 − tan x) − ( 1 + tan x)( 1 − tan x) ′ ( 1 − tan x)^2

sec^2 x( 1 − tan x) − ( 1 + tan x)( − sec^2 x) ( 1 − tan x)^2

2 sec^2 x ( 1 − tan x)^2

(II.10) f (x) =

tan x 1 − tan x

Solution:

f ′ (x) = ( tan x) ′ ( 1 − tan x) − tan x( 1 − tan x) ′ ( 1 − tan x)^2

sec^2 x( 1 − tan x) − tan x( − sec^2 x) ( 1 − tan x)^2

sec^2 x ( 1 − tan x)^2

Universitas

Sask atchewanensis

DEOETPAT-RIÆ

2003 Doug MacLean

Solution Set III

Back to Questions

(III.1) f (x) = sin ( sin x)

Solution: f ′ (x) = ( cos ( sin x))( sin x) ′^ = ( cos ( sin x))( cos x) = cos x cos ( sin x)

(III.2) f (x) = tan^2 ( cos x)

Solution: f ′ (x) = 2 tan ( cos x)( tan ( cos x)) ′^ = 2 tan ( cos x)( sec^2 ( cos x))( cos x) ′^ =

2 tan ( cos x)( sec^2 ( cos x))( − sin x) = −2 sin x tan ( cos x) sec 2 ( cos x)

(III.3) f (x) = cot ( tan^3 x)

Solution: f ′ (x) = − csc 2 ( tan^3 x)( tan^3 x) ′^ = − csc^2 ( tan^3 x)( 3 tan^2 x( tan x) ′ ) =

− csc 2 ( tan^3 x)( 3 tan^2 x( sec^2 x) = −3 tan^2 x sec 2 x csc 2 ( tan^3 x) < 0

(III.4) f (x) = csc ( cot 4 x)

Solution: f ′ (x) = − csc ( cot 4 x) cot ( cot 4 x)( cot 4 x) ′^ = − csc ( cot 4 x) cot ( cot 4 x)( 4 cot 3 x)( cot x) ′^ =

− csc ( cot 4 x) cot ( cot 4 x)( 4 cot 3 x)( − csc 2 x) = 4 cot 3 x csc 2 x csc ( cot 4 x) cot ( cot 4 x)

Universitas

Sask atchewanensis

DEOETPAT-RIÆ

2003 Doug MacLean

Back to Questions

(III.9) f (x) = tan ( cot ( tan x))

f ′ (x) = sec^2 ( cot ( tan x))( cot ( tan x)) ′^ = sec^2 ( cot ( tan x))( − csc^2 ( tan x))( tan x) ′^ =

sec 2 ( cot ( tan x))( − csc^2 ( tan x))( sec^2 x) = − sec 2 x csc 2 ( tan x) sec^2 ( cot ( tan x)) < 0

(III.10) f (x) = csc

1 + x^2

Solution:

f ′ (x) = − csc

1 + x^2

cot

1 + x^2

( 1 + x^2 ) −^1

= − csc

1 + x^2

cot

1 + x^2

( − 1 )( 1 + x^2 ) −^2 ( 2 x)

− 4 x ( 1 + x^2 )^2 csc

1 + x^2

cot

1 + x^2