# Want to confirm $(f^{-1})'(x)\neq(f')^{-1}(x)$

I'm just making sure, that when I have "the inverse of the derivative of $f(x)$", that it's notated as $$(f')^{-1}(x)$$ and not the opposing "the derivative of the inverse of $f(x)$" which I think would be $$(f^{-1})'(x)$$

• You are right, although you should ensure your reader doesn't mix up your inverse-notation with $\frac{1}{f'}$
– Gono
Commented Jun 6, 2017 at 8:09
• Hmm. I would write the former at $f'(x)^{-1}$, possibly adding more parentheses for clarity: $\bigl(f'(x)\bigr)^{-1}$. When you write $(f')^(-1)(x)$, one interpretation could be different: $(f')^{-1}$ would be the inverse of the derivative, itself a map. And then you evaluate it at $x$. Commented Jun 6, 2017 at 8:09
• Followup to my previous comment: For example: $f(x)=x^4$, $f'(x)=4x^3$, $(f')^{-1}(y)=(y/4)^{1/3}$. Not what you intended, I think. Commented Jun 6, 2017 at 8:14

You're correct. For example, let $f: (0,\infty)\to\mathbb R$ such that $f(x)=x^2$. Then: \begin{align}f'(x) &= 2x\\ f^{-1}(x) &= \sqrt x\\ (f^{-1})'(x) &= \frac{1}{2\sqrt x}\\ (f')^{-1}(x) &= \frac{1}{2x} \end{align} The correct formula is $$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$$
• The RHS would read as "he reciprocal of the derivative of $f$ evaluated at the inverse of $f$ at $x$". Commented Jun 6, 2017 at 8:22