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I'm very confused, and this is probably a stupid question. I want to calculate $ \frac{d}{dx} f^{-1}(g^{-1}(x))$. However, I get two seemingly different results taking two different approaches.

I. $\frac{d}{dx} f^{-1}(g^{-1}(x)) = (f^{-1})'(g^{-1}(x)) \times (g^{-1})'(x) = \frac{1}{f'(f^{-1}(g^{-1}(x)))} \times \frac{1}{g'(g^{-1}(x))}$

II. $\frac{d}{dx} f^{-1}(g^{-1}(x)) = \frac{d}{dx} (g \circ f )^{-1}(x) = \frac{1}{(g \circ f)' ((g \circ f)^{-1} (x))}.$

Now, $(g \circ f)'(z) = g(f(z))\times f'(z)$ and $(g \circ f)^{-1} (x) = f^{-1}(g^{-1}(x))$, hence $\frac{d}{dx} f^{-1}(g^{-1}(x)) = \frac{1}{g(f(f^{-1}(g^{-1}(x)))) \times f'(f^{-1} (g^{-1}(x)))} = \frac{1}{f'(f^{-1} (g^{-1}(x)))}\times\frac{1}{ x}$.

Which way (if any) is correct, and why is the other one incorrect? Thanks so much!

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Your first approach is fine. Concerning the second one, you wrote that $(g\circ f)'(z)=g\bigl(f(z)\bigr)\times f'(z)$, but it should be $(g\circ f)'(z)=g'\bigl(f(z)\bigr)\times f'(z)$.

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  • $\begingroup$ oh, stupid mistake... thank you! :) $\endgroup$ – user509037 Apr 17 at 22:22

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