Note that $g'(x)=\frac{1}{f'(g(x))}$, where $g=f^{-1}$.
- If $f(x)=x^3+2x-1$, find $g'(2)$.
Here is how my professor did it:
We notice that $f(1)=2$, so $g(2)=1$. Then
$g'(2)=\frac{1}{f'(g(2))}=\frac{1}{f'(1)}=\frac{1}{5}$.
However, they then proceed to do another problem differently.
- If $f(x)=\sqrt{2x+3}$, find $g'(x)$ at $x=2$.
They did it this way:
$f(2)=\sqrt{7}$, so $g(\sqrt{7})=2$. Then
$g'(\sqrt{7})=\frac{1}{f'(g(\sqrt{7}))}=\frac{1}{f'(2)}=\frac{1}{\frac{1}{\sqrt{2(2)+3}}}=\frac{1}{\frac{1}{\sqrt{7}}}=\sqrt{7}$.
Following the first method, shouldn't this be:
$f(x)=2 \iff \sqrt{2x+3}=2 \iff x=\frac{1}{2}$
$g'(2)=\frac{1}{f'(g(2))}=\frac{1}{f'(\frac{1}{2})}=...$.
The statement of the second problem is somewhat confusing as both $f$ and $g'$ are written in terms of the variable $x$, and then it is specified that $x=2$. However the most reasonable interpretation is $$\hbox{if $f(x)=\sqrt{2x+3}$, find $g'(2)$}\ .$$ Assuming this is what is meant, the solution is wrong and your correction is right.
Because of this possibility of confusion, IMHO the best way to solve this kind of problem is to use two different variables. I would write it this way: let $y=f(x)$. Then $x=g(y)$ and so we want $\frac{dx}{dy}$ when $y=2$. We have $$y=2\ \Leftrightarrow\ f(x)=2\ \Leftrightarrow\ \sqrt{2x+3}=2\ \Leftrightarrow\ x=\frac12$$ and so $$\frac{dx}{dy}=1\Big/\frac{dy}{dx}=1\Big/\frac{1}{\sqrt{2x+3}}=2\ .$$ In fact in this case, $\frac{dy}{dx}$ is very closely related to $y$, $$\frac{dy}{dx}=\frac1y\ ,$$ so you can take a short cut, $$\frac{dx}{dy}=1\Big/\frac{dy}{dx}=y=2\ .$$