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.

I wrote an answer to a question that was closed as a duplicate of this one. I thought I would add a different answer to this question.

Integer Case
For integer $n\ge0$, the Binomial Theorem says
$$
(x+h)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}h^k\tag1
$$
So
$$
\frac{(x+h)^n-x^n}h=\sum_{k=1}^n\binom{n}{k}x^{n-k}h^{k-1}\tag2
$$
Therefore, taking the limit, we get
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}x^n
&=\binom{n}{1}x^{n-1}\\[3pt]
&=nx^{n-1}\tag3
\end{align}
$$

Inverting
Suppose that
$$
x=y^m\tag4
$$
Taking the derivative of $(4)$ using $(3)$ and substituting $y=x^{\frac1m}$ yields
$$
1=my^{m-1}\frac{\mathrm{d}y}{\mathrm{d}x}=mx^{1-\frac1m}\frac{\mathrm{d}}{\mathrm{d}x}x^{\frac1m}\tag5
$$
Therefore,
$$
\frac{\mathrm{d}}{\mathrm{d}x}x^{\frac1m}=\frac1mx^{\frac1m-1}\tag6
$$

Rational Case
Applying the Chain Rule with $(3)$ and $(6)$ gives
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}x^{\frac nm}
&=n\left(x^{\frac1m}\right)^{n-1}\frac1mx^{\frac1m-1}\\[3pt]
&=\frac nmx^{\frac nm-1}\tag7
\end{align}
$$

Real Case
When a sequence of functions converges pointwise and their derivatives converge uniformly, the derivative of the limit equals the limit of the derivatives (see this question). Therefore, the full case for non-negative exponents follows by continuity.

Negative Case
Applying the Chain Rule with $(7)$ and $\frac{\mathrm{d}}{\mathrm{d}x}\frac1x=-\frac1{x^2}$, we get
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}x^{-s}
&=-\frac1{(x^s)^2}sx^{s-1}\\
&=-sx^{-s-1}\tag8
\end{align}
$$
