1. ## Differentiation proof

For a postive integers $n$

we have $( x^n)' = nx^{n-1}$

for a fraction $\frac{p}{q}$

we also have $(x^{\frac{p}{q}})' = \frac{p}{q} x^{\frac{p}{q} -1}$

I can complete the above proofs but don't know to prove for irrational numbers index $n$ ? Is $(x^{n})' = n x^{n-1}$ ?

Moreover , are $(x^{\pi} )' = \pi x^{\pi -1}$ and $(x^{e} )' = e x^{e -1}$ ?
Thank you !

2. Originally Posted by simplependulum
For a postive integers $n$

we have $( x^n)' = nx^{n-1}$

for a fraction $\frac{p}{q}$

we also have $(x^{\frac{p}{q}})' = \frac{p}{q} x^{\frac{p}{q} -1}$

I can complete the above proofs but don't know to prove for irrational numbers index $n$ ? Is $(x^{n})' = n x^{n-1}$ ?

Moreover , are $(x^{\pi} )' = \pi x^{\pi -1}$ and $(x^{e} )' = e x^{e -1}$ ?
Thank you !
Proof of the Power Rule

3. Originally Posted by simplependulum
For a postive integers $n$

we have $( x^n)' = nx^{n-1}$

for a fraction $\frac{p}{q}$

we also have $(x^{\frac{p}{q}})' = \frac{p}{q} x^{\frac{p}{q} -1}$

I can complete the above proofs but don't know to prove for irrational numbers index $n$ ? Is $(x^{n})' = n x^{n-1}$ ?

Moreover , are $(x^{\pi} )' = \pi x^{\pi -1}$ and $(x^{e} )' = e x^{e -1}$ ?
Thank you !
First we have: $x^a,ax^{a-1}\in C[-\infty,+\infty]$(a is the indepedent variable).
Let $a$ be a irrational number. Let $\{a_n\}$be a rational sequence with $\lim a_n=a$.
We get a sequence of functions $\{f_n=x^{a_n}\}$.
Now we are to prove that $\lim f'_n=(\lim f_n)'$, so that $(x^a)'=\lim a_nx^{a_n-1}=ax^{a-1}$.
There is a theorem saying that:
If 1: $f_i\in C^{1}[c,d]$,
2: $\{f'_n\}$uniformly converges at $g$,
3: $\{f_n\}$converges at at least one point $x_0$,
then $\lim f'_n=(\lim f_n)'$.
we know that for every x, $\lim x^{a_n}=x^{a}$,so 3 is true.
1 is clearly true.
Now we only have to show 2.Let $g(x)=ax^{a-1}$
let $b_n=\sup_{x\in [c,d]}\{|f'_n(x)-g(x)|\}$, it suffices to show that $\lim b_n=0$
Let $e_n=a_n-a$, then $\lim e_n=0$
so $a_nx^{a_{n-1}}-ax^{a-1}=e_nx^{a_{n-1}}+a(x^{a_{n-1}}-x^{a-1})$
We have $x\in [c,d],|x^{a_{n-1}}|.
so $\forall\epsilon>0,\exists N,\forall n>N,|e_n|<\epsilon,|x^{a_{n-1}}-x^{a-1}|<\epsilon\forall x\in [c,d]$,the second inequality is true since $x^{p-1}$is uniformly continous on $[c,d]$.
So $|a_nx^{a_{n-1}}-ax^{a-1}|,we are done..
So tough a proof...

4. Or, if you first prove that the derivative of ln(x) is 1/x, starting from $y= x^r$, take the ln of both sides: ln(y)= r ln(x).

Now, differentiating both sides with respect to x, $\frac{1}{y}\frac{dy}{dx}= \frac{r}{x}$. Multiply both sides by y: $\frac{dy}{dx}= \frac{r}{x}y= \frac{r}{x}x^r= r x^{r-1}$.

That's easier than ynj's proof but requires first proving that the derivative of ln(x) is 1/x as well as the chain rule.