Prove that $(x^{\alpha})'=\alpha x^{\alpha-1}$ use the fact that $x^{\alpha}=e^{\alpha log(x)}$
I am close but I feel like I am missing a step here:
$(e^{\alpha log(x)})'$
using the product rule and taking the deriviative times the inside we get:
outside: $e^{\alpha log(x)}=x^{\alpha}$ whose derivative is $
\alpha x^{\alpha-1}$
but then the derivative of the inside is $\frac{\alpha}{x}$
combining the two gives:
$\alpha x^{\alpha-1}\frac{\alpha}{x}$
which is not correct what am I doing wrong?
 A: The derivative of $e^{\alpha \log(x)}$ is actually $e^{\alpha \log(x)} \frac{d}{dx}(\alpha \log(x))$ from chain rule which is nothing more than $\frac{\alpha}{x} e^{\alpha \log(x)}$. Can you take it from here?
A: Must you use $x=e^{\ln x}$?  Personally I like using logarithmic differentiation instead:
\begin{align}
y&=x^\alpha\\
\ln y&= \ln x^\alpha\\
\ln y&=\alpha \ln x\\
\frac{d}{dx} (\ln y)&=\frac{d}{dx} \alpha\ln x\\
\frac1y\cdot\frac{dy}{dx}&=\alpha\cdot\frac1x\\
\frac{dy}{dx}&=\alpha\cdot\frac1x\cdot y\\
\frac{dy}{dx}&=\alpha\cdot\frac1x\cdot x^\alpha\\
\frac{dy}{dx}&=\alpha\cdot x^{\alpha-1}
\end{align}
A: By definition,
$$(x^\alpha)'=\lim_{h\to 0}\dfrac{(x+h)^\alpha-x^\alpha}{h} $$
If $\alpha \in{\mathbb N}$, then by the binomial theorem,
$$\dfrac{(x+h)^\alpha-x^\alpha}{h}=\alpha x^{\alpha-1}+\frac{\alpha(\alpha-1)}{2}hx^{\alpha-2}+\ ... +h^{\alpha-1}$$
As $h\to 0$, the terms with $h$ tends to $0$, hence
$$\forall{\alpha \in{\mathbb{N}}}, \; \lim_{h\to 0}\dfrac{(x+h)^\alpha-x^\alpha}{h}=\alpha x^{\alpha-1}$$
Since
$(x+h)^{-\alpha}-x^{-\alpha}= \dfrac{x^{\alpha}-(x+h)^{\alpha}}{(x+h)^{\alpha}x^{\alpha}}\;,$
$$(x^{-\alpha})'=\lim_{h\to 0}\dfrac{(x+h)^{-\alpha}-x^{-\alpha}}{h}=\lim_{h\to 0}\dfrac{x^{\alpha}-(x+h)^{\alpha}}{h(x+h)^{\alpha}x^{\alpha}}$$ 
Notice that $\dfrac{x^{\alpha}-(x+h)^{\alpha}}{h}=-(x^{\alpha})'$, and as $h \to 0,\; (x+h)^{\alpha} \to x^{\alpha}.$
Therefore,
$$\lim_{h\to 0}\dfrac{x^{\alpha}-(x+h)^{\alpha}}{h(x+h)^{\alpha}x^{\alpha}}=-\alpha x^{\alpha -1}\frac{1}{x^{2\alpha}}=-\alpha x^{-\alpha -1}$$
Let $\beta = -\alpha$, and we have 
$$(x^{\beta})'=\beta x^{\beta -1}$$
By the formula for $a^n-b^n$,
$$ \lim_{h\to 0} \dfrac{(x+h)^{\frac{1}{\alpha}}-x^{\frac{1}{\alpha}}}{h}=\lim_{h\to 0} \dfrac{1}{\sum_{k=0}^{\alpha-1} ((x+h)^{\frac{1}{\alpha}})^{\alpha-1-k} \;(x^{\frac{1}{\alpha}})^{k}}$$
Again, as $h \to 0$, $x+h \to x$, so
$${\sum_{k=0}^{\alpha-1} ((x+h)^{\frac{1}{\alpha}})^{\alpha-1-k} \;(x^{\frac{1}{\alpha}})^{k}}=\alpha x^{\frac{\alpha-1}{\alpha}}$$
Hence,
$$ \lim_{h\to 0} \dfrac{(x+h)^{\frac{1}{\alpha}}-x^{\frac{1}{\alpha}}}{h}=\frac{1}{\alpha} x^{\frac{1-\alpha}{\alpha}}$$
If we let $\gamma = \frac{1}{\alpha}$, we have
$$(x^{\gamma})'=\gamma x^{\gamma-1}$$
So we have 
$$\forall{\eta\in{\mathbb Z}\cup \{\frac{1}{a}|a\in\Bbb Z\} },\;(x^{\eta})'=\eta x^{\eta-1}$$
By chain rule,
$$(x^{\frac{\eta_1}{\eta_2}})'=\eta_1 x^{\frac{\eta_1 -1}{\eta_2}}\frac{1}{\eta_2}x^{\frac{1}{\eta_2}-1}=\frac{\eta_1}{\eta_2}x^{\frac{\eta_1}{\eta_2}-1}$$
Therefore,
$$\forall{\omega\in \mathbb Q}, (x^{\omega})'= \omega x^{\omega -1}$$
EDIT: A way using $x^{\alpha}=e^{\alpha ln(x)}$:
$$\lim_{h\to 0} \frac{e^{\alpha ln(x+h)}-e^{\alpha ln(x)}}{h}=\lim_{h\to 0}\frac{e^{\alpha ln(x)} (e^{\alpha ln(\frac{x+h}{x})}-1)}{h}$$
$$\lim_{h \to 0}\frac{1}{h}(e^{\alpha ln(\frac{x+h}{x})}-1)=\lim_{h\to 0}\frac{1}{h}((1+\frac{h}{x})^{\alpha}-1)$$
By the binomial expansion, we have 
$$\lim_{h\to 0}\frac{1}{h}((1+\frac{h}{x})^{\alpha}-1)=\frac{\alpha}{x}$$
Hence, $$(x^{\alpha})'=\frac{\alpha}{x}x^{\alpha}=\alpha x^{\alpha-1} \;\;\forall{\alpha\in {\mathbb Q}}$$
A: The chain rule says that $\frac{d(f(y))}{dx}=\frac{d(f(y))}{dy}\cdot \frac{dy}{dx}$
Many people see this rule as the $dy$s cancelling out. However, it is not advised to look at it like that.
In your case $f(y)$=$e^y$, where $y=alog(x)$.
So, $\frac{d(f(y))}{dy}=\frac{d(e^y)}{dy}=e^y=e^{alog(x)}=x^a$
And $\frac{dy}{dx}=\frac{d(alog(x))}{dx}=\frac{a}{x}$
Multiply these together to get $ax^{a-1}$
