Question

Calculate the derivative of the function.
g(x) = (2x^2 + x + 1)^−3
Step by step please. I understand you use the chain rule, but then it confuses me after that.

Answer 1

think of this as
\[f(g(x))\] with
\[g(x)=2x^2+x+1\] the "inside function" and
\[f(u)=u^{-3}\] the "outside function" so that
\[f(g(x))=(x^2+x+1)^{-3}\]

Answer 2

then by the chain rule
\[f(g(x))'=f'(g(x))\times g'(x)=-3(2x^2+x+1)^{-4}\times (2x+1)\]

Answer 3

or think of it like this
\[y=u^{-3}\]
\[u=2x^2+x+1\]
\[\frac{dy}{du}=-3u^{-4}\]
\[\frac{du}{dx}=2x+1\] and
\[\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}=-3u^{-4}\times (2x+1)=-3(2x^2+x+1)^{-4}\times (2x+1)\]

Answer 4

Thank you this was very helpful.

Answer 5

yw