Derivatives of Trigonometric Functions

Derivatives of trigonometric functions are unlike the derivatives of algebraic functions.

Derivatives of some trigonometric functions using the first principle method:

sin x, cos x, tan x

Derivative of Sin x

f(x) = sin x

f '(x) = lim h 0 f(x+h)-f(x) h  

f '(x) = lim h 0 sin(x+h)-sin(x) h    

       = lim h 0 2 cos( 2x+h 2 )sin( h 2 ) h

  Since, sin A -sin B = 2 cos(A+B / 2) sin(A-B / 2)

         = lim h 0 2 cos( 2x+h 2 )sin( h 2 ) h   

As h → 0, we have h/2 → 0

          = lim h 0 cos( 2x+h 2 ) . lim h/2 0 sin( h 2 ) h 2                 

= cos x × 1

fI(x) = cos x

Therefore, the derivative of sin x is cos x.

Derivative of Cos x

f(x) = cos x

     f '(x) = lim h 0 cos(x+h)-cosx  h      

= lim h 0 - 2 sin( 2x+h 2 )sin( h 2 ) h

Since cos A - cos B = -2 sin(A+B /2)sin(A-B /2)

as h → 0, we have h/2 → 0

= lim h 0 - sin( 2x+h 2 ) . lim h/2 0 sin( h 2 ) h 2

= (- sin x) × 1    (∴ lim x 0 sin x x = 1)

fI(x) = -sin x

Therefore, the derivative of cos x is - sin x.

Derivative of Tan x

f(x) = tan x

f '(x) = lim h 0 tan(x+h)- tanx h

= lim h 0 1 h [ sin(x+h) cos(x+h) - sin x cos x ]

= lim h 0 1 h [ sin(x+h).cos x - sin x. cos(x+h) cos(x+h) ]

= lim h 0 1 h [ sin(x+h-x) cos(x+h).cos x ]     [Since sin(A - B) = sin A cos B - cos A sin B]         

= lim h 0 1 h [ sin(h) cos(x+h).cos x ]   

= lim h 0 sin h h [   lim h 0   1 cos(x+h).cos x ]

= 1 × [1/(cos x cos x)]

= 1 × [1/(cos2 x)]

= sec2 x

The derivative of tan x is secant2 x.

Summary

Derivatives of trigonometric functions are unlike the derivatives of algebraic functions.

Derivatives of some trigonometric functions using the first principle method:

sin x, cos x, tan x

Derivative of Sin x

f(x) = sin x

f '(x) = lim h 0 f(x+h)-f(x) h  

f '(x) = lim h 0 sin(x+h)-sin(x) h    

       = lim h 0 2 cos( 2x+h 2 )sin( h 2 ) h

  Since, sin A -sin B = 2 cos(A+B / 2) sin(A-B / 2)

         = lim h 0 2 cos( 2x+h 2 )sin( h 2 ) h   

As h → 0, we have h/2 → 0

          = lim h 0 cos( 2x+h 2 ) . lim h/2 0 sin( h 2 ) h 2                 

= cos x × 1

fI(x) = cos x

Therefore, the derivative of sin x is cos x.

Derivative of Cos x

f(x) = cos x

     f '(x) = lim h 0 cos(x+h)-cosx  h      

= lim h 0 - 2 sin( 2x+h 2 )sin( h 2 ) h

Since cos A - cos B = -2 sin(A+B /2)sin(A-B /2)

as h → 0, we have h/2 → 0

= lim h 0 - sin( 2x+h 2 ) . lim h/2 0 sin( h 2 ) h 2

= (- sin x) × 1    (∴ lim x 0 sin x x = 1)

fI(x) = -sin x

Therefore, the derivative of cos x is - sin x.

Derivative of Tan x

f(x) = tan x

f '(x) = lim h 0 tan(x+h)- tanx h

= lim h 0 1 h [ sin(x+h) cos(x+h) - sin x cos x ]

= lim h 0 1 h [ sin(x+h).cos x - sin x. cos(x+h) cos(x+h) ]

= lim h 0 1 h [ sin(x+h-x) cos(x+h).cos x ]     [Since sin(A - B) = sin A cos B - cos A sin B]         

= lim h 0 1 h [ sin(h) cos(x+h).cos x ]   

= lim h 0 sin h h [   lim h 0   1 cos(x+h).cos x ]

= 1 × [1/(cos x cos x)]

= 1 × [1/(cos2 x)]

= sec2 x

The derivative of tan x is secant2 x.

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