Question

Find the derivative of the following functions from first principle. (i) x 3 – 27 (ii) ( x – 1) ( x – 2) (ii) (iv)

Answer 1

(I)

Let the given function be:

f( x )=( x 3 −27 )

From the definition, derivative of a function f( x ) over a point a can be written as:

f ' ( x )= lim h→0 f( x+h )−f( x ) h

On solving the value of f ' ( x ) , we get:

f ' ( x )= lim h→0 [ ( x+h ) 3 −27 ]−( x 3 −27 ) h = lim h→0 x 3 + h 3 +3 x 2 h+3 h 2 x−27− x 3 +27 h = lim h→0 h 3 +3 x 2 h+3 h 2 x h

Taking h common from numerator and denominator,

f ' ( x )= lim h→0 h 2 +3 x 2 +3 h 2 x = 0 2 +3 x 2 +3⋅ 0 2 ⋅x =3 x 2

Thus, the derivative of the function f( x )=( x 3 −27 ) is 3 x 2

(II)

Let the given function be:

f( x )=( x−1 )( x−2 )

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

On further simplification, we get

f ' ( x )= lim h→0 2hx+ h 2 −3h h = lim h→0 2x+h−3 (Taking h common)

On applying limits, we get

f ' ( x )=2x−3

Thus, the derivative of the function f( x )=( x−1 )( x−2 ) is ( 2x−3 ) .

(III)

Let the given function be:

f( x )= 1 x 2

f ' ( x )= lim h→0 f( x+h )−f( x ) h = lim h→0 1 ( x+h ) 2 − 1 x 2 h = lim h→0 1 h [ x 2 − ( x+h ) 2 ( x+h ) 2 x 2 ] = lim h→0 1 h [ x 2 − x 2 − h 2 −2hx ( x+h ) 2 x 2 ]

On further simplification, we get

f ' ( x )= lim h→0 1 h ( − h 2 −2hx ) ( x 2 ) ( x+h ) 2 = lim h→0 ( −h−2x ) ( x 2 )( x+h ) (Taking h common from numerator and denominator)

On applying limits, we get

f ' ( x )= ( 0−2x ) x 2 ( x+0 ) 2 = −2x x 2 ( x 2 ) = −2 x 3

Thus, the derivative of the given function is f( x )= 1 x 2 is −2 x 3 .

(IV)

Let the given function be:

f( x )= x+1 x−1

f ' ( x )= lim h→0 f( x+h )−f( x ) h = lim h→0 [ ( x+h+1 x+h−1 )−( x+1 x−1 ) ] h = lim h→0 1 h [ ( x−1 )( x+h+1 )−( x+1 )( x+h−1 ) ( x−1 )( x+h−1 ) ] = lim h→0 1 h [ x 2 +hx+x−x−h−1−( x 2 +hx+x−x+h−1 ) ( x−1 )( x+h−1 ) ]

On further simplification, we get

f ' ( x )= lim h→0 1 h ( −2h ) ( x−1 )( x+h−1 ) = lim h→0 ( −2 ) ( x−1 )( x+h−1 ) (Taking h common)

On applying limits, we get

f ' ( x )= −2 ( x−1 )( x+0−1 ) = −2 ( x−1 )( x−1 ) = −2 ( x−1 ) 2

Thus, the derivative of the function f( x )= x+1 x−1 is −2 ( x−1 ) 2 .