# Proof of derivative of invertible function

I know $y=f(x)$ is differentiable and invertible.

His inverse function is $x=g(y)$ which is also differentiable.

I have to prove that $g'(y)= 1/(f'(x))$.

I tried first with an baby example with $y=x^2$ and it turned out that this holds.

But I have no idea on how to prove this. Please provide some help!

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The correct equality is rather $g'(x)=1/f'(g(x))$.

Hint: Write $f \circ g= \operatorname{Id}$ and use the chain rule.

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Can you please be a bit more specific? – Sjoerd Smaal Feb 16 '13 at 13:49
@SjoerdSmaal: You only have to write $f(g(x))=x$ and then to differentiate the equality. – Seirios Feb 16 '13 at 13:52

Directly by definition, and putting $\,f(y)=x\Longleftrightarrow y= f^{-1}(x)=g(x)\,$:

$$g'(x_0):=\lim_{x\to x_0}\frac{g(x)-g(x_0)}{x-x_0}=\lim_{x\to x_0}\frac{y-y_0}{f(y)-f(y_0)}=\lim_{y\to y_0}\frac{1}{\frac{f(y)-f(y_0)}{y-y_0}}=\ldots$$

Now, you have to talk, perhaps, a little about the cases:

1) $\,f(y)=f(y_0)\,$ , a constant function in some neighborhood of $\,y_0\,$...but this is impossible (why?)

2) We have $\,x\to x_0\Longrightarrow g(x)=y\to g(x_0)=y_0\,$ (why?)

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