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diffrentiation by ab-initio method

can any help me out how to differentiate x^x by ab-initio method or first principle
lim⁡h→0(x+h)x+h−xxh=lim⁡h→0(x+h)x[lim⁡h→0(x+h)h−(xx+h)xh]=xx[lim⁡h→0(x+h)h−1h−lim⁡h→0(xx+h)x−1h]\displaystyle\begin{aligned}\lim_{h\to 0}\frac{(x+h)^{x+h}-x^x}{h}&=\lim_{h\to 0} (x+h)^x\left[\lim_{h\to 0}\frac{(x+h)^h-\left(\frac{x}{x+h}\right)^x}{h}\right]\\&=x^x\left[\lim_{h\to 0} \frac{(x+h)^h-1}{h}-\lim_{h\to 0} \frac{\left(\frac{x}{x+h}\right)^x-1}{h}\right] \end{aligned}

Expand both expressions binomially, noting that (xx+h)x=(1−hx+h)x\left(\frac{x}{x+h}\right)^x= \left(1-\frac{h}{x+h}\right)^x. The first expression will need a little more work afterwards.
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nasbah bm
OP
really thnx a lot :smile: for helpn me out
Reply 3
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Smaug123
15
Original post by Lord of the Flies
The first expression will need a little more work afterwards.

Starting from the beginning, I wrote out the first-principles definition of ddxxx\frac{d}{dx}x^x, stared at it and found it impossible. I thought it was a bit more "obvious what to do at each step" to:

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prove the chain rule

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prove that exp differentiates to itself

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prove the product rule

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prove the "inverse function theorem" as my course called it: ddxf−1(x)=1f′(f−1(x))\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}

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prove that log differentiates to 1/x

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put it all together


especially since each individual bit is standard bookwork.
(edited 11 years ago)
Original post by Smaug123
...


Well yes, but a binomial expansion is also standard textbook, no? As well as being substantially quicker!
Reply 5
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Smaug123
15
Original post by Lord of the Flies
Well yes, but a binomial expansion is also standard textbook, no? As well as being substantially quicker!

This is true, but embarrassingly enough it didn't occur to me to carry one out…

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