Inverse function derivative

Can you explain further as I see it, if:

$\displaystyle y = f^{-1}(x) \Leftrightarrow f(y) = x.$

Differentiate with respect to y:

$\displaystyle \frac{dx}{dy} = f^{'} (y) = f^{'} \Big( f^{-1} (x) \Big).$

Therefore:

$\displaystyle \frac{dy}{dx} = \frac{ d[f^{-1} (x)] }{dx} = \frac{1}{f^{'} \Big( f^{-1} (x) \Big)}.$

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Off topic: Is this to do with the "Inverse Function Theorem"?

So $\displaystyle \frac{d}{dx} f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}$

I understand this can be shown from the identity map $\displaystyle f \circ f^{-1}(x) = x$ and just differentiating and making $\displaystyle \frac{d}{dx} f^{-1}(x)$ the subject.

But this looked familiar when my lecture notes said it's the same as $\displaystyle \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$ if you let $\displaystyle y = f^{-1}(x)$

Now I'm probably being stupid but, wouldn't that make the above $\displaystyle \frac{dy}{dx} = \frac{1}{\frac{d}{dx}y}= \frac{1}{\frac{dy}{dx}}[br] [br]$ ?

How did you get the last line From what's above?

It might be, but I'm just reading the rules for differentiation.

I believe you are missing the last part of the equation, the F and exponent can't just disappear and should be included, so you have 2/3's of the equation. Apologies if I am incorrect and on the wrong track, its been a while since I had quadratic equations, Calc 1 & 2, etc.

I applied the reciprocal of dx/dy (so I did 1/(dx/dy) ). :-)

The equation is written properly. I'm not sure what you mean... the f^-1 is there.

Yes, it is correct in the last one you had there, but I was referring to the last line of your original...I was writing my response and by the time I had completed, you had made a correction to your initial post-which is the one (the first one) I was responding too. So now it is indeed correct.

I haven't edited my post, maybe it took while to load or something I don't know.

But I still don't understand how my working is wrong :/

it isn't correct as it would mean that dy/dx at some point is always equal to it's reciprocal.

The equation as stated by Simon is correct, and that is what I was noting-you quoted it in responding to him and asking about it-what he gave was the correct one as far as I can tell.

Oh ok, but what I want to know is how what I've written is incorrect (concerning the relationship between notation), because it seems like the answer was used to prove the original question.