How do you find the derivative of $x^{2/5}$ using the definition of the derivative?
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3Whatever $a$ is, the derivative of $x^a$ is $a x^{a-1}$ – Claude Leibovici Sep 15 '14 at 18:45
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2i think he is asking f(x+h)-f(x)/h method – avz2611 Sep 15 '14 at 18:47
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I posted a detailed evaluation (using the definition of the derivative) of the derivative of $x^{\frac{m}{n}},$ where $m$ and $n$ are nonzero integers (positive or negative), for my answer to the math StackExchange question Differentiation using first principles with rational powers. – Dave L. Renfro Sep 15 '14 at 19:12
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$$(x^{2/5})'\\ =\lim_{h\to0}\frac{(x+h)^{2/5}-x^{2/5}}h\\ =\lim_{h\to0}\frac{(x+h)^{2/5}-x^{2/5}}h\frac{(x+h)^{8/5}+(x+h)^{6/5}x^{2/5}+(x+h)^{4/5}x^{4/5}+(x+h)^{2/5}x^{2/5}+x^{8/6}}{(x+h)^{8/5}+(x+h)^{6/5}x^{2/5}+(x+h)^{4/5}x^{4/5}+(x+h)^{2/5}x^{2/5}+x^{8/5}}\\ =\lim_{h\to0}\frac{(x+h)^2-x^2}h\frac1{(x+h)^{8/5}+(x+h)^{6/5}x^{2/5}+(x+h)^{4/5}x^{4/5}+(x+h)^{2/5}x^{2/5}+x^{8/6}}\\=\frac{2x}{5x^{8/5}}\\ =\frac25x^{-3/5}.$$ The trick is to make a binomial like $a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$ appear, to get rid of the $2/5$ exponent.
It will work for all rational exponents $p/q$, yielding the difference of two $p^{th}$ powers at the numerator and $q$ terms in $x^{p/q}$ raised at the $(q-1)^{th}$ at the denominator: $$(x^{p/q})'=\frac{px^{p-1}}{qx^{(q-1)p/q}}=\frac pqx^{p/q-1}.$$
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This is interesting, how do you do this in general. I'm not sure I understand what you mean with the binominal, as the mentioned one does not appear in the formulas. – Alice Ryhl Sep 15 '14 at 18:57
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Indeed it is an artifice. It is better known for the square roots, like $$\sqrt a-\sqrt b=\frac{a-b}{\sqrt a+\sqrt b},$$ but the idea is the same. – Sep 15 '14 at 19:06