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    can any help me out how to differentiate x^x by ab-initio method or first principle
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    \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 \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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    really thnx a lot for helpn me out
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    (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 \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:
    • prove the chain rule
    • prove that exp differentiates to itself
    • prove the product rule
    • prove the "inverse function theorem" as my course called it: \frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}
    • prove that log differentiates to 1/x
    • put it all together

    especially since each individual bit is standard bookwork.
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    (Original post by Smaug123)
    ...
    Well yes, but a binomial expansion is also standard textbook, no? As well as being substantially quicker!
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    (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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