So, I was given the following problem as part of a homework assignment.

Suppose $f'(x) > 0$ in $(a,b)$. Prove that $f$ is strictly increasing in $(a,b)$, and let $g$ be its inverse function. Prove that $g$ is differentiable, and that $$g'(f(x)) = \frac{1}{f'(x)}$$

I have proven that $f$ is strictly increasing in $(a,b)$, and I could prove that $g'(f(x)) = 1/f'(x)$ if I could prove that $g$ is differentiable. The problem is that I am having trouble with a proof of that. Any advice?

Also, as a reference, this is exercise 5.2 from Baby Rudin.

  • $\begingroup$ Recall that by definition of an inverse, $g$ must satisfy $g(f(x))=x$. $\endgroup$ – Joshua Gray Dec 3 '15 at 0:34
  • $\begingroup$ Yes, but how does this show that $g$ is differentiable? $\endgroup$ – Birdman2246 Dec 3 '15 at 0:37
  • 3
    $\begingroup$ Honestly, from a purely geometric approach this is clear. The inverse is just a reflection over $y = x$, and so our tangent line at each point would just be the reciprocal. Since we have no zero derivative, no singularities arise upon reflecting. By analytic techniques, I would ask you this: The right hand side of our composition is certainly differentiable, and so is $f$. If $g$ were not differentiable, what would this imply? $\endgroup$ – Rellek Dec 3 '15 at 0:47
  • $\begingroup$ So you are proposing a proof by contradiction? I can't think of a single contradiction that would be implied from $g$ not being differentiable. $\endgroup$ – Birdman2246 Dec 3 '15 at 0:57
  • $\begingroup$ Honestly I kind of led you astray with the last sentence. I was thinking of a certain contrapositive but it was not related here. $\endgroup$ – Rellek Dec 3 '15 at 1:25

Recall that if $f$ is invertible, then it is bijective. So, there exists a unique $y$ such that $y = f(x)$ on the domain of $g$, which is seen to be $(f(a),f(b))$ since we have an increasing function. By definition of the derivative:

$$g'(y) = \lim_{z \rightarrow f(x)} \frac{g(z) - g(f(x))}{z-f(x)} = \lim_{z \rightarrow f(x)} \frac{g(z) - x}{z-f(x)}$$

Now, as we tend $z$ closer and closer to $f(x)$, eventually it will have to belong to $(f(a),f(b))$, which means we can find another $x_z$ such that $f(x_z) = z$. We now choose $z$ sufficiently close, and take advantage of this fact. We then have:

$$g'(y) = \lim_{z \rightarrow f(x)} \frac{g(f(x_z)) - x}{f(x_z)-f(x)} = \lim_{z \rightarrow f(x)} \frac{x_z - x}{f(x_z)-f(x)}$$

We see that this final limit tends to $\frac{1}{f'(x)}$.


We know that $g$ is the inverse of $f$ defined on $(f(a),f(b))$. This means that $g\circ f:(a,b)\rightarrow (a,b)$ is the identity function, that is $$g(f(x)) = x.$$

This function is clearly differentiable with derivative

$$\frac{d}{dx}(g\circ f)(x) = 1$$

for all $x\in (a,b)$. This is the same as saying that

$$\lim_{t\rightarrow x}\frac{g(f(t))-g(f(x))}{t-x}=1.$$

Now let $s,y\in (f(a),f(b))$ and consider the difference quotient


We know that there exists unique $t,x\in (a,b)$ such that $f(t) = s$ and $f(x) = y$ but then

$$\frac{g(s)-g(y)}{s-y} = \frac{g(f(t))-g(f(x))}{t-x}\cdot \frac{t-x}{f(t)-f(x)} =\frac{g(f(t))-g(f(x))}{t-x}\cdot \frac{1}{\frac{f(t)-f(x)}{t-x}}\rightarrow 1\cdot \frac{1}{f'(x)}$$

as $s\rightarrow y$.

Note that we can apply the change of variables without knowing anything of $g$ for denote

$$h(s) = \frac{g(s)-g(y)}{s-y}$$

then we know that $h(f(t))\rightarrow \frac{1}{f'(x)}$ as $t\rightarrow x$ (assuming that $y = f(x)$). Let $\varepsilon$ be given then we can find a $\delta>0$ such that $t\in (x-\delta,x+\delta)$ implies that $|h(f(t))-\frac{1}{f'(x)}|<\varepsilon$ but then if $s\in (f(x-\delta),f(x+\delta))$ then $|h(s)-\frac{1}{f'(x)}|<\varepsilon$.


Lemma: $g:U \rightarrow \mathbb{R}$ is differentiable at $x$ with derivative $A$ iff there exists a function $\phi:U \rightarrow \mathbb{R}$ continuous at $x$ such that $g(t)-g(x)=\phi(t)(t-x)$ for all $t \in U$ and $\phi(x)=A$.

The above lemma should be easy to prove. Apply it.


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