derivative with non integer (irrational) as power

Question

How do you prove the power rule can be applied for the derivative of $$x^\pi?$$Since the power is an irrational number, how to use the binomial theorem in this case ?

Answer 1

You can use the chain rule and the following trick :

$$x^n=e^{\ln(x)\cdot n}$$

Which has derivate

$$e^{\ln(x)\cdot n}\cdot \frac{n}{x}=n\cdot x^{n-1}$$

for every real $\ n\ $ and positive $\ x\ $

Answer 2

Let $y=x^\pi$

$$\ln y = \pi \ln x$$ $$\implies\ln y = \pi \ln x$$ $$\implies\frac{d\,y}{y} = \pi \frac{d \,x}{x}$$ $$\implies\frac{d\,y}{d\,x} = \pi \frac{y}{x} = \pi x^{\pi-1}$$

Coming back to your question the binomial theorem applies even for irrational indices

So you can us the binomial expansion to prove it as well.

Answer 3

As arthur mentioned in the comments, you first have to define what the symbol $x^y$ means when $x>0$, and $y \in \mathbb{R}$ (regardless of irrational or not). One typical approach is to first define the logarithm and exponential function, prove a bunch of their properties, and AFTER THAT DEFINE $x^y = e^{y \log(x)}$. Then you can prove that \begin{equation} \dfrac{d}{dx}(x^y) = y \cdot x^{y-1} \end{equation}

Notice the logic of how things progressed: first, define what the symbol $x^y$ is supposed to mean. Then, prove a bunch of properties which are relevant such as: \begin{align} x^y \cdot x^z = x^{y+z} \tag{*} \end{align} Finally, deduce the differentiation rule as a simple corollary. The toughest part of all this is DEFINING $x^y$. For more information about the details, I highly suggest reading Michael Spivak's book: Calculus, Chapter 18 which is on logarithm and exponential function.

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Anyway, here's a proof (all these equal signs need to be carefully justified, and for that you should read the book I suggested, or something else) \begin{align} \dfrac{d}{dx}(x^y) &= \dfrac{d}{dx}(e^{y \log(x)}) \tag{by definition} \\ &= y \cdot \dfrac{d(\log(x))}{dx} \cdot e^{y \log(x)} \tag{chain rule} \\ &= y \cdot \dfrac{1}{x} \cdot x^y \tag{property of $\log$ and $e$} \\ &= y \cdot x^{y-1} \tag{by (*)} \end{align}