Derivative of $x^a$ How to calculate the derivative of function $f$ such, that:
$f(x)=x^a$, where $x>0$ and $a\in \Bbb R$.
Do you know any formula for $(x+h)^a$?
 A: For $a$ an integer one can prove that $f'(x)=ax^{a-1}$ by induction. In the general case we can use the chain rule:
$$
f(x)=e^{a\log x}
$$
so
$$
f'(x)=e^{a\log x}\cdot a\cdot \frac{1}{x}=ax^{a-1}
$$
Of course $f$ is defined only for $x>0$ if $a$ is not supposed to be an integer, but an arbitrary real number.
A: Yes. You can show it very easily what it is starting from the definition. Let's make the example when $a$ is an integer (as a hint) (you know what the definition of the derivative is?).
$$
f'(y)=\lim_{x\to y} \frac{x^a-y^a}{x-y}=\frac{(x-y)(x^{a-1}+ax^{a-2}+...)}{x-y}=ax^{a-1}
$$
I skipped some steps, but I think you can try to reproduce them as an exercise. If is not clear let us know, and we can help you more. 
A: Try logarithmic differentiation:  if $f(x) = x^a$, then $\ln f(x) = a \ln x$, whence
$(1 / f(x)) f'(x) = a / x$.  Thus $f'(x) = af(x) / x = ax^a /x = ax^{a - 1}$, just as if $a$ were an integer.  
I can't, off the top of my head, quote a formula for $(x + h)^a$, but I'm pretty sure something along those lines is known.  Might be a challenge to do the algebra required in computing the derivative, to wit $((x + h)^a - x^a) / h$, when $a$ is not a postivie integer.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
