Formula:

Chain rule -

Where u and v are the functions of x.

(i) (2x + 3) (3x – 5)

Applying, Chain rule

Here, u = 2x + 3

V = 3x -5

(2x + 3) (3x – 5)

= (2x + 3)(3x^1-1+0) + (3x – 5)(2x^1-1+0)

= 6x + 9 + 6x -10

= 12x -1

(ii) x(1 + x)³

Applying, Chain rule

Here, u = x

V = (1 + x)³

x(1 + x)³

= x×3×(1 + x)² + (1 + x)³(1)

= (1 + x)²(3x+x+1)

= (1 + x)²(4x+1)

(iii) = (x^1/2 + x^-1)(x – x^-1/2 )

Applying, Chain rule

Here, u = (x^1/2 + x^-1)

V = (x – x^-1/2 )

(x^1/2 + x^-1)(x – x^-1/2 )

= (x^1/2 + x^-1)(x – x^-1/2 ) + (x – x^-1/2 )(x^1/2 + x^-1)

= (x^1/2 + x^-1)(1+ x^-3/2) + (x – x^-1/2 )(x^-1/2 – x^-2)

= x^1/2 + x^-1 + x^-1 + x^-5/2 + x^1/2 – x^-1 - x^-1 + x^-5/2

= x^1/2 + x^-5/2

(iv)

Differentiation of composite function can be done by

Here, f(g) = g² , g(x) =

= 2g×(1 + )

= 2( (1 + )

= 2(x + - + )

= 2(x + )

(v)

Differentiation of composite function can be done by

Here, f(g) = g³ , g(x) = x² -

= 3g²×(2x - )

= 3 (2x - )

= 3(2x³ - - + )

= 3(2x³ - + )

(vi) (2x² + 5x – 1) (x – 3)

Applying, Chain rule

Here, u = (2x² + 5x – 1)

V = (x – 3)

(2x² + 5x – 1) (x – 3)

(2x² + 5x – 1)(2x² + 5x – 1)

= (2x² + 5x – 1)×1 + (x – 3)(4x + 5)

= 2x² + 5x – 1 + 4x² -7x -15

= 6x² -2x -16