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How can we prove this formula?

Let $f$ and $f^{-1}$ are differentiable.

$$f(x_0)=y_0\quad \Leftrightarrow\quad f^{-1}(y_0)=x_0$$

Formula of derivative of inverse function is that:

$$\dfrac{d}{dx}\left(f^{-1}\right)(y_0)=\dfrac{1}{f'(x_0)}$$

But why? Where this formula comes from?

Idea 1:

Using the features of inverse functions.

We accualy know that function and its inverse fuction(If it exists) show symmetric property, like given below. enter image description here

Therefore, if we observe tangent on a point in $f$ like $A(x_0,y_0)$ (then in inverse function mean $A'(y_0,x_0)$) ,and I think, we can say that there is relationship between $f'(x_0)$ and $f^{-1}(y_0)$ ,but how? Here I couldn't complete.

Idea 2:

Definition of limit of derivative.

$$\lim\limits_{h\to 0}\dfrac{f(x_0+h)-f(x_0)}{h}=f'(x_0)$$

$$\lim\limits_{h\to 0}\dfrac{f^{-1}(y_0+h)-f^{-1}(y_0)}{h}=\dfrac{d}{dy}\left(f^{-1}\right)(y_0)$$

Here I couldn't see any hint to prove consequence, as well...

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    see also here https://en.wikipedia.org/wiki/Inverse_functions_and_differentiation – 2017-01-15
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    How can we prove this formula? – 2017-01-15
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    i hope this will help you http://www.math.jhu.edu/~jmb/note/invfnthm.pdf – 2017-01-15
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    If you already know the chain rule, you can differentiate $f^{-1}\circ f$ (whichs equals the identity function). – 2017-01-15
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    thank you all, Git Gud and Dr.Sonnhard Graubner, I solved it :):):) – 2017-01-15
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    And if you can suggest other methots, I would listen :) – 2017-01-15

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