Find the derivative of $g\circ f^{-1}$ where $g(x) = x^{3}-x$ and $f(x)= x^{3} + x$. Find the derivative of $g\circ f^{-1}$ where $g(x) = x^{3}-x$ and $f(x)= x^{3} + x$ at $x=2$.
My attempt:
\begin{align*}\frac{d(g\circ f^{-1})}{dx} &= g'(f^{-1}(x))((f^{-1})'(x))(f'(x))\\&= g'(f^{-1}(2))(f^{-1}(2))'(f'(2))\end{align*}
$$\{ (f^{-1})'(f(x))= 1/f'(x) => (f^{-1}(2))' = 1/f'(1) = 1/4 \}$$
\begin{align*} &= g'(1)(1/4)(13)\\
 &= 2.\frac{1}{4}.13\\
 &= 13/2\end{align*}
Is this right way to find the derivative.
Answer is given = $1/2$
 A: Say $h(x) = f^{-1}(x)$. Then you have
$$\frac{d}{dx}(g\circ h)(x) = g’(h(x)) h’(x).$$
Your error lies in having an extra $f’(x)$ at the end of that formula, for no apparent reason.
Now; what is $h’(x)$? You can use the Inverse Function Theorem: if $f(a)=b$, then
$$(f^{-1})’(b) = \frac{1}{f’(a)}.$$
Here note that $f(1)=2$. So
$$(f^{—1})’(2) = \frac{1}{f’(1)}.$$
You should be able to proceed from there.
A: Everything is correct, except that you applied the chain rule incorrectly in the first step; applying the chain rule should instead give
$$(g\circ f^{-1})'(x)=g'(f^{-1}(x))\cdot (f^{-1})'(x)$$
where there is no factor of $f'(x)$ - this is just taking the rule for any two functions and applying it to one that happens to be the inverse of some other function. This is why your answer is off by a factor of $13$: you multiplied by $f'(2)=13$ unnecessarily.
It's worth noting that you can do this a bit more clearly with differentials; if you say that $y=g(f^{-1}(x))$ you can define $z=f^{-1}(x)$ and note that $f(z)=x$ and $y=g(z)$. Then you get
$$dx=f'(z)\,dz$$
$$dy=g'(z)\,dz$$
and hence
$$\frac{dy}{dx}=\frac{g'(z)}{f'(z)}$$
and since you can calculate $z=1$ when $x=2$, this gives the answer in a straightforwards manner.
