Power Rule Tip

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calculusexpert

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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)
 
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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: \(\displaystyle \frac{d }{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\))
 
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]
 
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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.
 
The Highlander said:
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: \(\displaystyle \frac{d }{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\))
Click to expand...
Shared a "TIP" only instead of any Question etc. Its for just a simple learning tip
 
The Highlander said:
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.
Click to expand...
The OP had posted another "tip" recently and Ted responded to the post saying that this is ok.
 
calculusexpert said:
Shared a "TIP" only instead of any Question etc. Its for just a simple learning tip
Click to expand...
Steven G said:
The OP had posted another "tip" recently and Ted responded to the post saying that this is ok.
Click to expand...
That's fine, I wasn't complaining. I was just a little bit puzzled. 🤷‍♂️

It took me a couple of minutes to figure out exactly what I (finally decided) the string "d/dx[x^(-2)] = -2x^(-3) and d/dx[√x] = d/dx[x^(1/2)] = (1/2)x^(-1/2)" actually meant, which is why I felt re-posting it (with LaTex formatting) might be useful? 🤔

And, after doing so, I couldn't see that it offered much beyond the standard 'formula' we teach for differentiation:-
[math]\left(\frac{d }{dx}ax^n=nax^{(n-1)}\right)[/math]
However, as I say, I have no complaints about it (other than its formatting); the content is indisputable. 🙂
 
The important point in the tip is that the power rule works not only for positive integers, which is where we typically first use it, but for any real number. That includes not only negative numbers and fractions, but even irrational numbers!

I have see trick questions, intended to make a student think carefully, that ask you to differentiate these functions:
[math]x^\pi\\\pi^x\\\pi^{\pi}[/math]
They're all easy, but easy to get wrong!
 
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