Power Rule Tip

[calculusexpert]

Power Rule Tip
The power rule works for all real numbers n, including negative and fractional exponents! For example: d/dx[x^(-2)] = -2x^(-3) and d/dx[√x] = d/dx[x^(1/2)] = (1/2)x^(-1/2)

 

[The Highlander]

I found your layout rather confusing.

Did you mean to write:- [math]\frac{d }{dx}x^{-2}=-2x^{-3}\text{ and }\frac{d }{dx}\sqrt{x}=\frac{d }{dx}x^{\frac{1}{2}}=\frac{1}{2}x^{-\frac{1}{2}}\>?[/math]
Can you explain what prompted you to offer this "Tip"?

It's not a question (seeking any help) and seems to have come 'out of the blue' with no apparent request from anyone looking for it.

(Personally, I would just write: \(\displaystyle \frac{d }{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\))

 

[fresh_42]

You can prove it for any exponent with the help of the chain rule and the derivative of the natural logarithm:
[math]\begin{array}{lll} \log x^\alpha&=\alpha\log x\\[10pt] \dfrac{d}{dx} \log\left(x^\alpha\right)&=\dfrac{1}{x^\alpha}\cdot \left(x^\alpha\right)'=\alpha\cdot \dfrac{d}{dx}\log x=\dfrac{\alpha}{x}\\[16pt] \left(x^\alpha\right)'&=\dfrac{\alpha}{x}\cdot x^\alpha=\alpha \,x^{\alpha-1} \end{array}[/math]

 

[fresh_42]

I just noticed that the derivative of the logarithm also follows from the chain rule, together with the definition of the exponential function as solution to [imath] y'=y \wedge y(0)=1.[/imath]

[math]\begin{array}{lll} x&=\exp(\log x)\\[10pt] 1&=\exp(\log x)\cdot (\log x)'\\[16pt] (\log x)'&=\dfrac{1}{\exp(\log x)}=\dfrac{1}{x} \end{array}[/math]
Now I wonder whether [imath] \left(x^\alpha\right)'=\alpha\,x^{ \alpha-1} [/imath] can be proven using only the chain rule, i.e., without using the exponential function in one way or another.