# FIND THE DERIVATIVE OF TRIGONOMETRIC FUNCTIONS

To find the derivative of trig functions, we may use the rules given below

d (sin x) / dx = cos x

d (cos x) / dx = -sin x

d (tan x) / dx = sec2 x

d (sec x) / dx = sec x tan x

d (cot x) / dx = -cosec2 x

d (cosec x) / dx = -cosec x cot x

Problem 1 :

√x tan x

Solution :

=﻿ d/dx [√x tan x]

Problem 2 :

√(x tan x)

Solution :

= d/dx [√(x tan x)]

Problem 3 :

sin 2x + cos 3x

Solution :

= d/dx [sin 2x + cos 3x]

= cos 2x · d/dx (2x( + (- sin 3x) · d/dx [3x]

= cos 2x (2) - sin 3x (3)

dy/dx = 2 cos 2x - 3 sin 3x

Problem 4 :

sin (4θ + π/2)

Solution :

= sin (4θ + π/2)

y = cos

dy/dθ = d/dθ [cos (4θ)]

dy/dθ = (-sin 4θ) · d/dθ [4θ]

dy/dθ = -4 sin (4θ)

Problem 5 :

tan θ + sec θ

Solution :

= d/dx [tan θ + sec θ]

= d/dθ [tan θ] + d/dθ [sec θ]

= sec² (θ) + sec (θ) tan (θ)

Factoring sec θ

= sec (θ) [tan (θ) + sec (θ)]

Problem 6 :

sin 2x cos 3x

Solution :

= d/dx [sin 2x cos 3x]

Since two differentiable functions are multiplied, we can find the derivative using product rule.

u = sin 2x ==> u' = cos 2x (2) ==> u' = 2 cos 2x

v = cos 3x ==> v' = - sin 3x (3) ==> u' = -3 sin 3x

= 2 cos (2x) cos (3x) - 3 sin (2x) sin (3x)

Problem 7 :

sin (2 cos 3x)

Solution :

= d/dx [sin (2 cos 3x)]

= cos (2 cos (3x)) · d/dx [2 cos (3x)]

= cos (2 cos (3x)) · 2 (-sin 3x) (3)

= - 6 cos (2 cos (3x)) (sin 3x)

= -6 sin (3x) cos (2 cos (3x))

Problem 8 :

sec³ x4

Solution :

= d/dx [sec³ x4]

= 3 sec² (x4) sec (x4) tan (x4) d/dx [x4]

= 3 tan (x4) 4 sec3 (x4) x3

= 12x³ sec³ (x4) tan (x4)

Problem 9 :

arctan (3x - 5)

Solution :

y = tan-1(3x - 5)

Problem 10 :

arcsin (3x - 5)

Solution :

y = sin-1(3x - 5)

Problem 11 :

2x+cosx

Solution :

y = 2x+cosx

d(ax)/dx = ax ln a

= d/dx [2x+cosx]

= ln (2) · 2x+cosx · d/dx [x + cos x]

= ln (2) · 2x+cosx · d/dx [x] + d/dx [cos x]

= ln (2) · 2x+cosx · (1 - sin x)

Problem 12 :

y = x log x

Solution :

y = x log x

Since two differentiable functions are multiplied, we use the product rule.

u = x and v = log x

u' = 1 and v' = 1/x

dy/dx = x (1/x) + log x (1)

dy/dx = 1 + log x

Problem 13 :

y = (sin x)²

Solution :

= d/dx [(sin x)²]

dy/dx = 2 sin (x) d/dx [sin (x)]

dy/dx = 2 sin (x) cos (x)

dy/dx = sin 2x

Problem 14 :

y = sin 2x - cos 4x

Solution :

= d/dx [sin 2x - cos 4x]

= d/dx [sin 2x] - d/dx [cos 4x]

= cos 2x d/dx [2x] - (-sin 4x) d/dx [4x]

= cos (2x) (2) + sin 4x (4)

= 2 cos (2x) + 4 sin 4x

= 4 sin 4x + 2 cos 2x

Problem 15:

y = 4x sin x

Solution :

= d/dx [4x sin(x)]

= 4 d/dx [x sin(x)]

u = x and v = sin x

u' = 1 and v' = cos x

= 4 [ x cos x + sin x (1) ]

= 4x cos x + 4 sin x

Problem 16 :

y = log(x² + 1)

Solution :

= d/dx [log(x² + 1)]

= 1/(x² + 1) · d/dx [x² + 1]

= 2x + 0 / (x² + 1)

= 2x / (x² + 1)

Problem 17:

y = ex sin x

Solution:

= d/dx [ex sin x]

= d/dx [ex] sin (x) + ex · d/dx [sin (x)]

= ex sin (x) + ex cos (x)

= ex [sin (x) + cos (x)]

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