Derivative of $x^n$ for $n < 0$

Asked Dec 01 '22 at 06:49

Active Dec 01 '22 at 06:59

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I know that you can prove the derivative of $x^n$ where $n > 0$ using the bionomial theorem etc. Now I have looked online and couldn't find out but I'd like to ask here, is the a proof for the derivative of $x^n$ where $n<0$?

edited Dec 01 '22 at 06:59

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asked Dec 01 '22 at 06:49

Nav Bhatthal

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MathJax tip: curly braces {} are used for grouping, and are not rendered. If you want to put a superscript containing more than a single symbol, group those symbols in curly braces. For example, $x^{-n}$ renders as $x^{-n}$. – Theo Bendit Dec 01 '22 at 06:54
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Does this answer your question? – Naitik Mundra Dec 01 '22 at 07:02
Thank you both! – Nav Bhatthal Dec 01 '22 at 07:05
Note that you can directly prove for negative integer $n$ by noting that $(x+h)^n-x^n=[x^{-n}-(x+h)^{-n}]/x^n(x+h)^n$, and then applying the binomial theorem to the numerator. – Glen O Dec 01 '22 at 07:17
Once I've applied the theorem to the numerator how do I clear up the denominator? – Nav Bhatthal Dec 01 '22 at 07:45
Nevermind I've done it – Nav Bhatthal Dec 01 '22 at 07:51

Derivative of $x^n$ for $n < 0$

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