Question

Find the derivative of the following functions from first principle: (i) – x (ii) (– x ) –1 (iii) sin ( x + 1) (iv)

Answer 1

(i)

Let the given function be f( x )=−x.

By first principle, the derivative of given function is,

f ′ ( x )= lim h→0 [ −( x+h )−( −x ) ] h = lim h→0 ( −x−h+x ) h = lim h→0 −h h =−1

(ii)

Let the given function be f( x )= ( −x ) −1 .

On simplifying, the function becomes,

f( x )= −1 x

By first principle, the derivative of given function is,

f ′ ( x )= lim h→0 1 h ( −1 x+h −( −1 x ) ) = lim h→0 1 h ( −x+x+h x( x+h ) ) = lim h→0 1 h ( h x( x+h ) ) = 1 x 2

(iii)

Let the given function be f( x )=sin( x+1 ).

By first principle, the derivative of given function is,

f ′ ( x )= lim h→0 [ sin( x+h+1 )−sin( x+1 ) h ] = lim h→0 1 h [ 2cos( x+h+1+x+1 2 )sin( x+h+1−x−1 2 ) ] = lim h→0 1 h [ 2cos( 2x+h+2 2 )sin( h 2 ) ] = lim h→0 [ cos( 2x+h+2 2 ) sin( h 2 ) h 2 ]

Simplify further,

f ′ ( x )= lim h→0 cos( 2x+h+2 2 ) lim h 2 →0 sin( h 2 ) h 2 =cos( 2x+0+2 2 )( 1 ) =cos( x+1 )

(iv)

Let the given function be f( x )=cos( x− π 8 )

By first principle, the derivative of given function is,

f ′ ( x )= lim h→0 cos( x+h− π 8 )−cos( x− π 8 ) h = lim h→0 1 h [ −2sin ( x+h− π 8 +x− π 8 ) 2 sin ( x+h− π 8 −x+ π 8 ) 2 ] = lim h→0 1 h [ −2sin( 2x+h− π 4 2 )sin( h 2 ) ] = lim h→0 [ −sin( 2x+h− π 4 2 ) sin( h 2 ) h 2 ]

Further simplify,

f ′ ( x )= lim h→0 [ −sin( 2x+h− π 4 2 ) ] lim h 2 →0 sin h 2 h 2 =−sin( 2x+0− π 4 2 )( 1 ) =−sin( x− π 8 )