Find derivative of inverse of function $y=2x^3-6x$ and calculate it's value at $x=-2$.
My Approach:
We know that $(f^{-1}(f(x)))=x$
Taking derivative both side
$(f^{-1}(f(x)))' \cdot f'(x)=1$
$(f^{-1}(f(x)))'=\frac{1}{f'(x)}=\dfrac{1}{6x^2-6}$
We want to find $f^{-1}(-2)$ so we must have $f(x)=-2$ i.e. $\quad$$2x^3-6x=-2$
$\implies$ $2x^3-6x+2=0$
I can't find any integer root from here.
But given answer is $\frac{1}{18}$
I know other method to solve this problem but can we solve using my approach used above?
Best Answer
Denote the inverse of $f$ as $g(y)=f^{-1}(y)$
As you wrote, it holds $g[f(x)] = x$. Taking derivative both sides and applying chain rule yields
$$ g'[f(x)] f'(x) = 1$$
Thus $$ g'[f(x)] = \frac{1}{6(x^2-1)} $$ Evaluated at $x=-2$, the RHS is 1/18 as requested.