1. ## Differentiation proof

For a postive integers $\displaystyle n$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.