Question

Find the derivative pf ((x^2-1)^5-x)^3

Answer 1

So you have:
\[\frac{d}{dx}((x^2-1)^5-x)^3\]
So you have 2 chain rules. One for (x^2-1)^5 and the other for (...)^3; So we have:
\[\frac{d}{dx}f(g(h(x)))=f'(g(h(x))g'(h(x))h'(x)\]
Or practically:
\[(3)((x^2-1)^5-x)^2* \frac{d}{dx}((x^2-1)^5-x)\]
\[=3((x^2-1)^5-x)^2*(5)((x^2-1)^4-1)*\frac{d}{dx}(x^2-1)=\]
\[15((x^2-1)^5-x)((x^2-1)^4-1)(2x)=30x((x^2-1)^5-x)((x^2-1)^4-1)\]