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

• December 16th 2010, 11:42 AM
Diamondlance
Stewart's Proof that Inverses of Differentiable Functions are Differentiable
The following is paraphrased from Stewart's Calculus text.

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

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

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

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

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

...

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

The starred step is the one I'm having difficulty with understanding fully. I understand the continuity argument that as $x\rightarrow a$ we have $y\rightarrow b$, 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 $\lim_{x\rightarrow a}$ to $\lim_{y\rightarrow b}$ (although an $\epsilon - \delta$ argument seems to work in the opposite direction--that $\lim_{y\rightarrow b}h(x)=L$ would imply that $\lim_{x\rightarrow a}h(x)=L$). It seems that the claim in Stewart's argument also has to do with the fact that $f$ and $g$ are one-to-one. Any thoughts?
• December 16th 2010, 11:58 AM
Plato
Quote:

Originally Posted by Diamondlance
Since $f$ is differentiable, it is continuous, so $g$ is continuous by earlier theorem. Thus as $x\rightarrow a$ we have $y\rightarrow b$.

That seems to explain it does it not?
• December 16th 2010, 12:11 PM
snowtea
Think about it this way.

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

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

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

The rest of the step is just algebraic manipulation.
• December 16th 2010, 12:20 PM
Diamondlance
Snowtea, that makes sense, and helps to alleviate the problem I was having.

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

$\lim_{x\rightarrow 2}x=2$ whereas $\lim_{x^2\rightarrow 4}x$ doesn't make sense since $x$ is not a function of $x^2$ on $\mathbb{R}$. However, since $x$ is a continuous function of $y$ here, the switch is justified.
• December 16th 2010, 01:33 PM
Diamondlance
So I suppose the more rigorous argument I was searching for would go as follows:

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

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

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

where $\delta_2$ is small enough so that

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

$|\frac{g(x)-g(a)}{x-a}-g'(a)|<\epsilon$ by the assumed limit.
• December 16th 2010, 02:01 PM
snowtea
I don't think it is good notation for the limit to write $\displaystyle{\lim_{g(x)\rightarrow g(a)}\frac{g(x)-g(a)}{x-a}}$.
It's better to write $\displaystyle{\lim_{y\rightarrow b}\frac{g(f(y))-g(a)}{f(y)-a}}$

we can rewrite
$\displaystyle{\frac{g(f(y))-g(a)}{f(y)-a}} = \frac{y-b}{f(y)-f(b)}}}$ by definition.
• December 16th 2010, 02:36 PM
Drexel28
Or we cuould just note that $f(g(x))=x$ and so by the chain rule we get $f'(g(x))g'(x)=1$. Dividing both sides by $f'(g(x))$ (since it's non-zero) gives the answer.
• December 16th 2010, 05:13 PM
Diamondlance
I don't think so, since the differentiability of $g(x)$ is not assumed.
• December 17th 2010, 03:25 AM
Plato
Quote:

Originally Posted by Diamondlance
I don't think so, since the differentiability of $g(x)$ is not assumed.

Actually that is a well known theorem.
If $f$ is a one-to-one differentiable function in an open interval $\mathcal{I}$ then $f^{-1}$ exists on $\mathcal{I}$ and is differentiable there.
• December 17th 2010, 08:13 AM
Diamondlance
Absolutely, but that's basically what we're trying to prove.