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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...

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3 Answers

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you have $$f(x) = -1 + \frac1{x^2 + 1}, \quad f(1) = -\frac 12, \quad f'(1) = -\frac14 $$ so the inverse function $g$ has $$g(-1/2) = 1, \quad g'(-1/2) = \frac1{f'(1)} = -4. $$

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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.

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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)}$

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