Finding the derivative of an inverse function. Let $f(x) = (-x^2)/(x^2+1)$. If g(x) is the inverse function of f(x) and f(1)=-1/2, what us g'(-1/2)?
Can someone explain how to do the above problem as I am not even sure where to start. Would I have to find the inverse of f(x) first or...
 A: you have $$f(x) = -1 + \frac1{x^2 + 1}, \quad f(1) = -\frac 12, \quad f'(1) = -\frac14  $$  so the inverse function $g$ has $$g(-1/2) = 1, \quad g'(-1/2) = \frac1{f'(1)} = -4. $$
A: Write:
$$y = - \frac {x^2}{x^2 + 1}$$
or, if you'd like,
$$y(x^2 + 1) = -x^2$$
whatever you find easier to work with. In any event, the derivative of the inverse of $y = f(x)$ is $\dfrac {\mathrm dx}{\mathrm dy} = (f^{-1})'(y)$
So you can implicitly differentiate with respect to $y$ and then plug in the points under consideration.
You might discover the general formula:
$$\dfrac {\mathrm dx}{\mathrm dy} = \frac 1 {\frac {\mathrm dy}{\mathrm dx}}$$
which might be multi valued or not exist, depending on the function you're considering. In general, you'll need an $x$ and a $y$ if $f$ is not invertible , and as always, bad things will happen if you try to divide by zero.
A: use the chain rule - since g is the inverse of f we have
$g(f(x))= x$
chain rule gives
$\frac{d}{dx}g(f(x)) = f'(x) g'(f(x)) = 1$
so
$f'(1) g'(f(1)) = f'(1) g'(-\frac 12)=1$
$ g'(-\frac 12)=\frac{1}{f'(1)}$
