# Inverse function theorem - derivative has not full rank

I am a little bit confused... Suppose that the derivative of a function $$f$$ has not full rank at a point $$a$$.

However, this does not imply that there doesn't exist an inverse function. We can only say that we are not allowed to apply the inverse function theorem which would guarantee the existence of an inverse function with nice properties, right?

• yes, that's right. An example is $f(x) = x^3$. An inverse function will, however, not be differentiable. – Thomas Jan 12 at 15:08
• @Thomas, thank you, I haven't thought about this simple example. Is there an intuition why these inverse functions cannot be differentiable? – Philipp Jan 12 at 15:15
• well, $f\circ f^{-1}(x) = x$ is for sure differentiable. If $f$ is in $x$ and $f^{-1}$ were in $f(x)$, too, write down what this implies for $d(f^{-1})$ at $f(x)$. (For simplicity I just made this claim for functions, i.e. in one dimension). – Thomas Jan 12 at 15:47
• @Thomas, I am not sure if I have understood your idea right. Let be $f: \mathbb{R}\supseteq M \to N \subseteq \mathbb{R}$ invertible on a neighborhood of $a$ where $df(a)=0$ and $f(a)=b$. If we assume that $f^{-1}$ is differentiable then we have $$df^{-1}(b)=lim_{y\to b}\frac{f^{-1}(y)-f^{-1}(b)}{y-b}=lim_{f(x)\to f(a)}\frac{x-a}{f(x)-f(a)},$$where $x \in M$ and $y \in N$. This looks like the reciprocal differential quotient of $f$ at point $a$ which would not be defined as $df(a)=0$. However, we take the limit $lim_{f(x)\to f(a)}$ instead of $lim_{x \to a}$. Or is this approach wrong? – Philipp Jan 12 at 20:37
• Differentiate the identity (apply the chain rule) to $f\circ f^{-1} (x) = x$ and try to figure out what it means for the derivative of the inverse function. – Thomas Jan 13 at 6:08