Proving the Derivative Rule for Rational Exponents

[jostpuur]

How do you prove the derivative

<br /> Dx^r = rx^{r-1}<br />

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

 

[jostpuur]

I know how to prove

<br /> f(x) = x^{n/m}\quad\implies\quad f&#039;(x) = \frac{n}{m}x^{n/m\;-1}<br />

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

<br /> 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}<br />

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

<br /> (f(x))^m = x^n<br />

and then solving f&#039;(x).

 

[gel]

<br /> D(\lim_{q\to r} x^q}) \underset{?}{=} \lim_{q\to r}(Dx^q)<br />

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&#039;_n converges uniformly towards something. Then f is differentiable and f&#039;_n\to f&#039;.

?

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&#039;_n converges uniformly towards something. Then f is differentiable and f&#039;_n\to f&#039;.

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.

<br /> f_n(x+h)=f_n(x) + h f_n^\prime(x+a_n)<br />

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

<br /> f(x+h)=f(x) + h g(x+a)<br />

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.

<br /> f_n(x+h)=f_n(x) + h f_n^\prime(x+a_n)<br />

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&#039;_n converges uniformyl to g"? But is there any reason to assume that \lim_{n\to\infty}a_n exists?

<br /> f(x+h)=f(x) + h g(x+a)<br />

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

 

[gel]

You mean "if f&#039;_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

<br /> \inf_{|a|&lt;|h|}f_n^\prime(x+a)\le(f_n(x+h)-f_n(x))/h\le \sup_{|a|&lt;|h|}f_n^\prime(x+a)<br />

then take limits as n->infinity

<br /> \inf_{|a|&lt;|h|}g(x+a)\le(f(x+h)-f(x))/h\le \sup_{|a|&lt;|h|}g(x+a)<br />

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