How to differentiate y=(x+2)^2(3x-1)^3y=(x+2)2(3x1)3 , find dy/dxdydx as a product of its factor ?

2 Answers

dy/dx=(x+2)(15x+16)(3x-1)^2dydx=(x+2)(15x+16)(3x1)2

Explanation:

Given that

y=(x+2)^2(3x-1)^3y=(x+2)2(3x1)3

differentiating w.r.t. xx as follows

dy/dx=d/dx((x+2)^2(3x-1)^3)dydx=ddx((x+2)2(3x1)3)

=(x+2)^2d/dx(3x-1)^3+(3x-1)^3d/dx(x+2)^2=(x+2)2ddx(3x1)3+(3x1)3ddx(x+2)2

=(x+2)^2(3(3x-1)^2(3))+(3x-1)^3(2(x+2))=(x+2)2(3(3x1)2(3))+(3x1)3(2(x+2))

=9(x+2)^2(3x-1)^2+2(x+2)(3x-1)^3=9(x+2)2(3x1)2+2(x+2)(3x1)3

=(x+2)(3x-1)^2(9x+18+6x-2)=(x+2)(3x1)2(9x+18+6x2)

=(x+2)(3x-1)^2(15x+16)=(x+2)(3x1)2(15x+16)

=(x+2)(15x+16)(3x-1)^2=(x+2)(15x+16)(3x1)2

Jul 17, 2018

dy/dx=2(x+2)(3x-1)^3+9(x+2)^2(3x-1)^2dydx=2(x+2)(3x1)3+9(x+2)2(3x1)2

Explanation:

By the product rule

(uv)'=u'v+uv'

and the chain rule

(f(g(x)))'=f'(g(x))*g'(x) we get

dy/dx=2(x+2)(3x-1)^3+(x+2)^2*3(3x-1)^2*3
this is equal to

dy/dx=2(x+2)(3x-1)^3+9(x+2)^2(3x-1)^2