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Can we find the derivative of the inverse of a function implicitly by finding the derivative of the original function?

For example lets say I have $f(x) = e^x$ and I want to find the derivative of the inverse function $f^{-1}(x) = ln(x)$, without actually finding the inverse function itself and then taking the derivative of it (which would be $\frac{1}{x}$)


For further clarification of what I'm trying to ask, this is the standard way of finding the derivative of the inverse of a given function $f(x) = e^x$

$$\text{Given} \ \ f : x \to e^x \ \ \forall x \in \mathbb{R}$$ $$f(x) = e^x$$ $$\implies f^{-1}(x) = ln(x)$$ $$\frac{df^{-1}}{dx} = \frac{1}{x}$$

But what if I wanted to find the derivative of the inverse function, without having to first find the inverse function from the given function, and then taking the derivative of that.

Question : So given some function $f$, I want to know if it's possible to find $\frac{df^{-1}}{dx}$ without first finding $f^{-1}$ and then taking the derivative of it. In other words can you implicitly find $\frac{df^{-1}}{dx}$ given a function $f$?


I ask this question, as there would be functions in which finding the inverse functions may prove difficult or impossible (correct me if I'm wrong here), and this way of finding the derivative of the inverse of the function, may or may not be useful in those cases.

I also haven't stated whether function $f$ (which I'm trying to implicitly find the derivative of its inverse) is injective, bijective or surjective, I have done this on purpose as I'm sure the answer would differ depending on whether $f$ is injective, bijective or surjective.

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Let $y=f^{-1}(x) \iff x = f(y)$. Then leveraging the Chain Rule

$$\frac{d}{dx}[x] = \frac{d}{dx}[f(y)] \implies 1 = f'(y) \cdot \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{1}{f'(y)} = \frac{1}{f'(f^{-1}(x))}$$

Hence $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$

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  • $\begingroup$ @ Stefan4024 But would that not just produce $\frac{1}{x} \ \ \forall \ f \ \ $? $\endgroup$ – Perturbative Apr 30 '16 at 14:28
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    $\begingroup$ I think you left out the $'$ symbol to denote the derivative of $f$ in the denominator, as Evinda has done in his/her answer below. $\endgroup$ – Perturbative Apr 30 '16 at 14:50
  • $\begingroup$ @Perturbative No because we have derivative $\endgroup$ – Stefan4024 Apr 30 '16 at 23:51
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It holds that $$f(f^{-1}(x))=x$$

So: $$(f(f^{-1}(x)))'=1 \Rightarrow f'(f^{-1}(x)) (f^{-1}(x))'=1 \Rightarrow (f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}$$

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