Proving the Derivative Rule for Rational Exponents

[jostpuur]

How do you prove the derivative

$$Dx^r = rx^{r-1}$$

for $r\in\mathbb{R}\backslash\mathbb{Q}$?

 

[jostpuur]

I know how to prove

$$f(x) = x^{n/m}\quad\implies\quad f'(x) = \frac{n}{m}x^{n/m\;-1}$$

for $n,m\in\mathbb{Z}$, but since the derivative mapping is not continuous, you cannot easily commute the limit and derivative like this

$$D x^r = D(\lim_{q\to r} x^q}) \underset{?}{=} \lim_{q\to r}(Dx^q) = \lim_{q\to r} qx^{q-1} = rx^{r-1}$$

For rational exponents you can carry out the proof by taking the derivative of the both sides of

$$(f(x))^m = x^n$$

and then solving $f'(x)$.

 

[gel]

$$D(\lim_{q\to r} x^q}) \underset{?}{=} \lim_{q\to r}(Dx^q)$$

That's fine, just as long as you can show that the limit on the right hand side converges uniformly (on an open interval containing x).

 

[HallsofIvy]

If you can differentiate exponentials and logarithms, then you can write $x^r= e^{r ln(x)}$. By the chain rule, the derivative of that is $(r/x)e^{r ln(x)}= (r/x)x^r= r x^{r-1}$. Of course the derivatives of e^x and ln(x) can be defined without reference to x^r.

 

[jostpuur]

That's fine, just as long as you can show that the limit on the right hand side converges uniformly (on an open interval containing x).

Is this true:

Let $I\subset \mathbb{R}$ be some open interval. Let $f_n:I\to\mathbb{R}$ be a sequence of differentiable functions, such that $f_n\to f$ uniformly, where $f$ is some function, and such that $f'_n$ converges uniformly towards something. Then $f$ is differentiable and $f'_n\to f'$.

?

If you can differentiate exponentials and logarithms, then you can write $x^r= e^{r ln(x)}$. By the chain rule, the derivative of that is $(r/x)e^{r ln(x)}= (r/x)x^r= r x^{r-1}$. Of course the derivatives of e^x and ln(x) can be defined without reference to x^r.

Surprising trick!

 

[gel]

Is this true:

Let $I\subset \mathbb{R}$ be some open interval. Let $f_n:I\to\mathbb{R}$ be a sequence of differentiable functions, such that $f_n\to f$ uniformly, where $f$ is some function, and such that $f'_n$ converges uniformly towards something. Then $f$ is differentiable and $f'_n\to f'$.

What HallsOfIvy said is probably the best way to approach this problem, but yes this is true. One way to prove it is to use the mean value theorem.

$$f_n(x+h)=f_n(x) + h f_n^\prime(x+a_n)$$

where |a_n|<|h|. If $f_n$ converges uniformly to g and a is a limit point of a_n then

$$f(x+h)=f(x) + h g(x+a)$$

Assuming g is continuous and using |a|<=|h| gives $f^\prime(x)=g(x)$.

 

[jostpuur]

What HallsOfIvy said is probably the best way to approach this problem, but yes this is true. One way to prove it is to use the mean value theorem.

$$f_n(x+h)=f_n(x) + h f_n^\prime(x+a_n)$$

where |a_n|<|h|. If $f_n$ converges uniformly to g and a is a limit point of a_n then

You mean "if $f'_n$ converges uniformyl to g"? But is there any reason to assume that $\lim_{n\to\infty}a_n$ exists?

$$f(x+h)=f(x) + h g(x+a)$$

Assuming g is continuous and using |a|<=|h| gives $f^\prime(x)=g(x)$.

 

[gel]

You mean "if $f'_n$ converges uniformyl to g"? But is there any reason to assume that $\lim_{n\to\infty}a_n$ exists?

Yes, that's what I should have said. You don't need to assume that a_n converges, as it is always possible to pass to a subsequence such that this is true. Alternatively use

$$\inf_{|a|<|h|}f_n^\prime(x+a)\le(f_n(x+h)-f_n(x))/h\le \sup_{|a|<|h|}f_n^\prime(x+a)$$

then take limits as n->infinity

$$\inf_{|a|<|h|}g(x+a)\le(f(x+h)-f(x))/h\le \sup_{|a|<|h|}g(x+a)$$

and letting h->0 gives $f^\prime(x)=g(x)$.