How do you find the derivative of g(x)=(2x^2+x+1)^-3g(x)=(2x2+x+1)3?

1 Answer
Apr 13, 2018

-3(4x+1)(2x^2+2x+1)^-33(4x+1)(2x2+2x+1)3

Explanation:

We use the chain rule, which states that,

dy/dx=dy/(du)*(du)/dxdydx=dydududx

Let u=2x^2+x+1,:.(du)/dx=4x+1.

Now, y=u^-3,:.dy/(du)=-3u^-4.

And so,

dy/dx=-3u^-4(4x+1)

Now, substitute back u=2x^2+x+1 to get:

dy/dx=-3(2x^2+x+1)^-3(4x+1)

I'd clean this up and rearrange it into:

=-3(4x+1)(2x^2+2x+1)^-3