Question is in the title, what is $\frac{d}{dx}[f^{-1}(x)]$ ? What I've done: Inverted $f(x) = 2x-4$ into $f^{-1}(x)=\frac{x+4}{2}$ and took the derivative of it. $\frac{d}{dx}[\frac{x+4}{2}] = \frac{1}{2}$, is this correct..? I didn't seem to use $x=2$, thus worried I've done something wrong. If I'm wrong, what's the correct answer? and what knowledge gap do I need to fill?
Trying to learn to tackle these types of question: Would a similar question: $\frac{d}{dx} f^{-1}(x)$ when $x=0$, given $f(x)=\frac{x}{1-x}$ and $f^{-1}(0)=0$ be Answer: $1$?
To answer your second question, if $f(x)=\frac{x}{1-x}$, then $f^{-1}(x)=\frac{x}{1+x}$. To compute $(f^{-1})'(0)$, you have two options: you can compute $(f^{-1})'(x)$ using the quotient rule, and then plug in $x=0$; or you can compute $f'(x)$ (again using the quotient rule), and then use the formula $$ (f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))} \, . $$ Plugging in $x=0$ gives the desired result. In this case, the first option looks like less work, but it is important to be aware of the formula for $(f^{-1})'(x)$, as it is applicable in a variety of circumstances.