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#