Answer :

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)(3x1-1+0) + (3x – 5)(2x1-1+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) = (x1/2 + x-1)(x – x-1/2 )


Applying, Chain rule


Here, u = (x1/2 + x-1)


V = (x – x-1/2 )


(x1/2 + x-1)(x – x-1/2 )


= (x1/2 + x-1)(x – x-1/2 ) + (x – x-1/2 )(x1/2 + x-1)


= (x1/2 + x-1)(1+ x-3/2) + (x – x-1/2 )(x-1/2 – x-2)


= x1/2 + x-1 + x-1 + x-5/2 + x1/2 – x-1 - x-1 + x-5/2


= x1/2 + 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


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