I have the following question here.
Let $F$ be a one-one-function such that $F=f \circ g$. If $g(1)=2$,$g'(1)=3$,$f(2)=3$,$f'(2)=4$, find $(F^{-1})'(3)$.
I'm not really sure how to approach this.
I know that $(F^{-1})'(3)=\frac{1}{F'(F^{-1}(3))}$ but I don't know the value of the inverse of the composition at $3$.
The only other thing I know is that $F'=f'(g(x))g'(x)$ but I'm not even sure if that's helpful here. Any help would be greatly appreciated thanks!
Since $F(x)=f\circ g(x)$, we have $x=F^{-1}\circ f\circ g(x)$. Differentiating both sides with respect to $x$,
$$1=\left(F^{-1}\right)'(f\circ g(x))f'(g(x))g'(x)$$
and when $x=1$,
$$1=\left(F^{-1}\right)'(f\circ g(1))f'(g(1))g'(1)=\left(F^{-1}\right)'(f(2))f'(2)3=\left(F{-1}\right)'(3)\cdot 4\cdot 3.$$