differentiate using first principle: x^2/3

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Solution

The trick is to use the difference of cubes formula
(a - b)(a² + ab + b²) = a³ - b³.
$\frac{d}{dx} x^{\frac{2}{3}} = \lim_{h \to 0} \frac{{(x + h)}^{\frac{2}{3}} - x^{\frac{2}{3}}}{h} = \lim_{h \to 0} \frac{{(x + h)}^{\frac{2}{3}} - x^{\frac{2}{3}}}{h} \times \frac{[{(x + h)}^{\frac{4}{3}} + x^{\frac{2}{3}} {(x + h)}^{\frac{2}{3}} + x^{\frac{4}{3}}]}{[{(x + h)}^{\frac{4}{3}} + x^{\frac{2}{3}} {(x + h)}^{\frac{2}{3}} + x^{\frac{4}{3}}]} = \lim_{h \to 0} \frac{{(x + h)}^{2} - x^{2}}{h [{(x + h)}^{\frac{4}{3}} + x^{\frac{2}{3}} {(x + h)}^{\frac{2}{3}} + x^{\frac{4}{3}}]} = \lim_{h \to 0} \frac{x^{2} + h^{2} + 2 xh - x^{2}}{h [{(x + h)}^{\frac{4}{3}} + x^{\frac{2}{3}} {(x + h)}^{\frac{2}{3}} + x^{\frac{4}{3}}]} = \lim_{h \to 0} \frac{h + 2 x}{[{(x + h)}^{\frac{4}{3}} + x^{\frac{2}{3}} {(x + h)}^{\frac{2}{3}} + x^{\frac{4}{3}}]} = \frac{2 x}{3 x^{\frac{4}{3}}} = \frac{2}{3} x^{\frac{- 1}{3}}$

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