# Thread: Stewart's Proof that Inverses of Differentiable Functions are Differentiable

1. ## Stewart's Proof that Inverses of Differentiable Functions are Differentiable

The following is paraphrased from Stewart's Calculus text.

Theorem: If $\displaystyle f$ is a one-to-one differentiable function with inverse function $\displaystyle g$ and $\displaystyle f'(g(a)) \ne 0$, then the inverse function is differentiable at $\displaystyle a$ and $\displaystyle g'(a)=\frac{1}{f'(g(a))}.$

Proof: We have $\displaystyle g'(a)=\displaystyle{\lim_{x\rightarrow a}\frac{g(x)-g(a)}{x-a}}$. Since $\displaystyle f$ and $\displaystyle g$ are inverses we have $\displaystyle g(x)=y$ iff $\displaystyle f(y)=x$ and $\displaystyle g(a)=b$ iff $\displaystyle f(b)=a$.

Since $\displaystyle f$ is differentiable, it is continuous, so $\displaystyle g$ is continuous by earlier theorem. Thus as $\displaystyle x\rightarrow a$ we have $\displaystyle g(x)\rightarrow g(a)$, or $\displaystyle y\rightarrow b$. Therefore:

$\displaystyle g'(a)=\displaystyle{\lim_{x\rightarrow a}\frac{g(x)-g(a)}{x-a}}$

**$\displaystyle =\displaystyle{\lim_{y\rightarrow b}\frac{y-b}{f(y)-f(b)}}$

...

$\displaystyle =\frac{1}{f'(g(a))}$

The starred step is the one I'm having difficulty with understanding fully. I understand the continuity argument that as $\displaystyle x\rightarrow a$ we have $\displaystyle y\rightarrow b$, but it seems like a bit of a leap, at least I can't think of any definitions or theorems to apply, to use that to change $\displaystyle \lim_{x\rightarrow a}$ to $\displaystyle \lim_{y\rightarrow b}$ (although an $\displaystyle \epsilon - \delta$ argument seems to work in the opposite direction--that $\displaystyle \lim_{y\rightarrow b}h(x)=L$ would imply that $\displaystyle \lim_{x\rightarrow a}h(x)=L$). It seems that the claim in Stewart's argument also has to do with the fact that $\displaystyle f$ and $\displaystyle g$ are one-to-one. Any thoughts?

2. Originally Posted by Diamondlance
Since $\displaystyle f$ is differentiable, it is continuous, so $\displaystyle g$ is continuous by earlier theorem. Thus as $\displaystyle x\rightarrow a$ we have $\displaystyle y\rightarrow b$.
That seems to explain it does it not?

3. Think about it this way.

Let $\displaystyle h(x) = \displaystyle{\frac{g(x)-g(a)}{x-a}}$.

So $\displaystyle g'(a) = \displaystyle{\lim_{x\rightarrow a}h(x)}$, and since $\displaystyle f$ is continuous and has $\displaystyle f'(b)\neq 0$:

$\displaystyle g'(a) = \displaystyle{\lim_{y\rightarrow b}h(f(y))}$

The rest of the step is just algebraic manipulation.

4. Snowtea, that makes sense, and helps to alleviate the problem I was having.

What I was thinking was that, for example, though $\displaystyle x\rightarrow 2$ implies $\displaystyle x^2\rightarrow 4$,

$\displaystyle \lim_{x\rightarrow 2}x=2$ whereas $\displaystyle \lim_{x^2\rightarrow 4}x$ doesn't make sense since $\displaystyle x$ is not a function of $\displaystyle x^2$ on $\displaystyle \mathbb{R}$. However, since $\displaystyle x$ is a continuous function of $\displaystyle y$ here, the switch is justified.

5. So I suppose the more rigorous argument I was searching for would go as follows:

Given one to one continuous function $\displaystyle g$ such that $\displaystyle \displaystyle{\lim_{g(x)\rightarrow g(a)}\frac{g(x)-g(a)}{x-a}}$ exists and is equal to $\displaystyle g'(a)$, we claim that $\displaystyle \displaystyle\lim_{x\rightarrow a}\frac{g(x)-g(a)}{x-a}}=g'(a)$. Given $\displaystyle \epsilon >0$, choose $\displaystyle \delta_1,\delta_2>0$ such that

$\displaystyle 0<|x-a|<\delta_1$

implies that $\displaystyle 0<|g(x)-g(a)|<\delta_2$ by continuity and injectivity of $\displaystyle g$,

where $\displaystyle \delta_2$ is small enough so that

$\displaystyle 0<|g(x)-g(a)|<\delta_2$ implies that

$\displaystyle |\frac{g(x)-g(a)}{x-a}-g'(a)|<\epsilon$ by the assumed limit.

6. I don't think it is good notation for the limit to write $\displaystyle \displaystyle{\lim_{g(x)\rightarrow g(a)}\frac{g(x)-g(a)}{x-a}}$.
It's better to write $\displaystyle \displaystyle{\lim_{y\rightarrow b}\frac{g(f(y))-g(a)}{f(y)-a}}$

we can rewrite
$\displaystyle \displaystyle{\frac{g(f(y))-g(a)}{f(y)-a}} = \frac{y-b}{f(y)-f(b)}}}$ by definition.

7. Or we cuould just note that $\displaystyle f(g(x))=x$ and so by the chain rule we get $\displaystyle f'(g(x))g'(x)=1$. Dividing both sides by $\displaystyle f'(g(x))$ (since it's non-zero) gives the answer.

8. I don't think so, since the differentiability of $\displaystyle g(x)$ is not assumed.

9. Originally Posted by Diamondlance
I don't think so, since the differentiability of $\displaystyle g(x)$ is not assumed.
Actually that is a well known theorem.
If $\displaystyle f$ is a one-to-one differentiable function in an open interval $\displaystyle \mathcal{I}$ then $\displaystyle f^{-1}$ exists on $\displaystyle \mathcal{I}$ and is differentiable there.

10. Absolutely, but that's basically what we're trying to prove.