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

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Answer 1 · 2018-04-13T04:44:32

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

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:

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

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