Prove that the derivative of $x^w$ is $w x^{w-1}$ for real $w$ Can anyone give a proof of the derivative of this type of function? Specifically showing that

$\dfrac{d(x^w)}{dx} = wx^{w-1}$ for a real $w$?

I tried to use the Taylor series expansion for $(x+dx)^w$ and got the correct result. However, the proof of the Taylor series requires knowledge of the derivative of these functions. So this is essentially circular reasoning. I know that the same series is also given by the binomial expansion, but that's not entirely satisfactory either, because where's the proof that the binomial expansion works for all reals (isn't it only apparent for integers)? So far all of the arguments I've come across involve circular reasoning.
I was thinking of showing that the binomial expansion is true for all reals using some form of proof by induction e.g. something like this. http://www.math.ucsd.edu/~benchow/BinomialTheorem.pdf
I'm really not sure.

We can use a method called logarithmic differentiation:
\begin{align}
y&=x^n\\
\log y&=\log x^n\\
\log y&=n\log x\\
\frac{d}{dx}\left(\log y\right)&=\frac{d}{dx}\left(n\log x\right)\\
\frac{y'}{y}&=\frac{n}{x}\\
y'&=\frac{n}{x}(y)\\
y'&=\frac{n}{x}(x^n)\\
y'&=nx^{n-1}
\end{align}
This uses the fact that $$\frac{d}{dx}\left(\log x\right)=\frac1x.$$

To see why this works for the part of the domain $x<0$ (since we are working with logarithms), we can consider that $$x=(-1)(-x)\implies x^n=(-1)^n(-x)^n$$ and we arrive at the same result, using the definition of the complex logarithm:
\begin{align}
y&=(-1)^n(-x)^n\\
\log y&=\log(-1)^n+\log(-x)^n\\
\log y&=\log(-1)^n+\log(|-x|)^n+i\pi\\
\end{align}
Here, on the right side of the equation, the first and last terms are constants, which yield a derivative of $0$.
