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Trigonometric Functions Derivatives Worksheet

Trigonometric Functions Derivatives Worksheet
  • Page 1
 1.  
Find Dxsec 9x.
a.
tan 9x sec x
b.
9 2 tan x sec 9x
c.
9 2 tan 9x sec 9x
d.
tan 9x sec 9x


Solution:

Dxsec 9x = Dx(sec 9x)12

= 12(sec 9x)-12Dx(sec 9x)
[Use the chain rule.]

= 12(sec 9x)-12(sec 9x tan 9x) 9

= 9 / 2 tan 9x sec 9x


Correct answer : (3)
 2.  
If t = sin 4x cos 4x, then find dtdx.
a.
sin 8x
b.
cos 8x
c.
cos x
d.
4cos 8x


Solution:

t = (sin 4x)(cos 4x) = 1 / 2 (sin 8x)
[Use sinθcosθ
= (1 / 2)sin2θ.]

Hence, dtdx = 1 / 2(8cos 8x) = 4cos 8x
[Use the chain rule.]


Correct answer : (4)
 3.  
If y = 8cos x21 + sin x, then find dydx.
a.
- 1+21sin x21+sin x
b.
- 8(1+21sin x)(21+sin x)2
c.
- 8(1+21sin x)21+sin x
d.
- 1+21sin x(21+sin x)2


Solution:

dydx = (21+sin x)Dx (8cos x) - 8cos xDx (21+sin x)(21+sin x)2
[Use the Quotient Rule.]

= (21 + sin x)(- 8sin x) - 8cos x(cos x)(21+sin x)2

= - 168sin x-8sin2 x-8cos2 x(21+sin x)2

= - 8(1+21sin x)(21+sin x)2
[Use sin² x + cos² x = 1.]

dydx = - 8(1+21sin x)(21+sin x)2


Correct answer : (2)
 4.  
If f (t) = (18 + 2sin t)52, then find f ′(t).
a.
5 (18 + 2sin t)32
b.
(18 + 2sin t)32 (cos t)
c.
5 (18 + 2sin t)32 (cos t)
d.
52(18 + 2sin t)32


Solution:

f ′(t) = 52(18 + 2sin t)32Dt(18 + 2sin t)
[Use the chain rule.]

= 52(18 + 2sin t)32 (2cos t)

= 5 (18 + 2sin t)32 (cos t)


Correct answer : (3)
 5.  
Find ddx[7sec (sin 3x)].
a.
7(sec (sin 3x) tan(sin 3x)) cos 3x
b.
(sec (sin 3x) tan(sin 3x)) cos 3x
c.
21(sec (sin 3x) tan(sin 3x)) cos 3x
d.
21(sec (sin 3x) tan(sin 3x)) cos x


Solution:

ddx[7sec (sin 3x)] = 7sec (sin 3x) tan (sin 3x)[Dx sin 3x]
[Use the Chain Rule.]

= 21(sec (sin 3x) tan(sin 3x)) cos 3x


Correct answer : (3)
 6.  
Find Dx(sin2 4x sin4 6x).
a.
4sin 8x sin4 6x + 24sin2 4x sin3 6x cos 6x
b.
4sin 8x sin4 6x + 24sin2 4x sin3 6x
c.
4sin 8x sin4 6x + 6sin2 4x sin3 6x cos 6x
d.
4sin 8x sin4 + 24sin2 4x sin3 6x cos 6x


Solution:

Dx(sin2 4x sin4 6x)

= [Dx (sin2 4x)](sin4 6x) + (sin2 4x)[Dx (sin4 6x)]
[Use the product rule.]

= (2 sin 4x) (Dx (sin 4x)) (sin4 6x) + (sin2 4x) (4sin3 6x)(Dx (sin 6x))

= (2sin 4x)(4 cos 4x) (sin4 6x) + (sin2 4x) (4sin3 6x)(6 cos 6x)

= 8sin 4x cos 4x sin4 6x + 24 sin2 4x sin3 6x cos 6x

= 4sin 8x sin4 6x + 24 sin2 4x sin3 6x cos 6x
[Use 2 sin θ cos θ = sin 2θ.]


Correct answer : (1)
 7.  
Differentiate f(x) = tan x72.
a.
x72 (sec2x72)
b.
x52 (sec2x72)
c.
(sec2x72)
d.
7x522 (sec2x72)


Solution:

f ′(x) = ddx (tanx72)

= (sec2 x72)ddx (x72)
[Use the chain rule.]

= (sec2 x72) 7 / 2 (x52)

= 7x522 (sec2x72)


Correct answer : (4)
 8.  
If y = cos2(4x + 34)72, then find dydx.
a.
- 14 (4x+34)52 [sin 2(4x+34)72]
b.
- [sin 2(4x+34)52]
c.
- (4x+34)52
d.
- 4 (4x+34)52 [sin 2(4x+34)72]


Solution:

dydx = 2[cos(4x+34)72] Dx [cos (4x+34)72]
[Use the chain rule.]

= [2cos (4x+34)72] [- sin(4x+34)72]Dx(4x+34)72
[Use the chain rule again.]

= - 2[sin (4x+34)72] [cos(4x+34)72] (72)(4x+34)52(4)

= - 14 (4x+34)52 [sin 2(4x+34)72]
[Use 2 sin θ cos θ = sin 2θ.]


Correct answer : (1)
 9.  
If y = sin 7x cos 9x, then find dydx.
a.
- 9 sin 7x sin 9x + 7 cos 7x cos 9x
b.
- 7sin 7x sin 9x + 9 cos 7x cos 9x
c.
- sin 7x sin 9x + cos 7x cos 9x
d.
9 sin 7x sin 9x - 7 cos 7x cos 9x


Solution:

dydx =ddx(sin 7x cos 9x) = sin 7xddx(cos 9x) + cos 9xddx(sin 7x)
[Use the Product Rule.]

= sin 7x [(- sin 9x)9] + cos 9x [(cos 7x)7]

= - 9 sin 7x sin 9x + 7 cos 7x cos 9x


Correct answer : (1)
 10.  
If y = 5cos2 x+sin2 3x, then find dydx.
a.
52cos2 x+sin2 3x (sin 2x-3sin 6xcos2 x+sin2 3x)
b.
does not exist
c.
52cos2 x+sin2 3x
d.
sin 2x-3sin 6xcos2 x+sin2 3x


Solution:

dydx = cos2 x+sin2 3xDx (5)-Dx (cos2 x+sin2 3x)cos2 x+sin2 3x
[Use the quotient rule.]

= - 52cos2 x+sin2 3x2 cosx(- sinx)+2sin 3x(cos 3x)(3)cos2 x+sin2 3x

= - 52cos2 x+sin2 3x(3sin 6x-sin 2xcos2 x+sin2 3x)
[Use 2sin θ cos θ = sin2 θ.]

= 52cos2 x+sin2 3x(sin 2x - 3sin 6xcos2 x+sin2 3x)


Correct answer : (1)

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