Question

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

Answer 1

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

Explanation:

We use the chain rule, which states that,

`dy/dx=dy/(du)*(du)/dx`

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`