Question

Find $\frac{dy}{dx}$, if $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$.

Answer 1

Given $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}},a$ is constant

$y^{\frac{2}{3}}=a^{\frac{2}{3}}-x^{\frac{2}{3}}$

$y={(a^{\frac{2}{3}}-x^{\frac{2}{3}})}^{\frac{3}{2}}$

Differentiating on both sides w.r.t $x$

$\frac{dy}{dx}=-\frac{3}{2}{(a^{\frac{2}{3}}-x^{\frac{2}{3}})}^{\frac{3}{2}-1}\frac{d}{dx}(a^{\frac{2}{3}}-x^{\frac{2}{3}})$

$=\frac{3}{2}\left(\sqrt{a^{\frac{2}{3}}-x^{\frac{2}{3}}}\right)\left(\frac{2}{3}\right)x^{(\frac{2}{3}-1)}$

$=-\frac{\sqrt{a^{\frac{2}{3}}-x^{\frac{2}{3}}}}{x^{\frac{1}{3}}}$

$\therefore \frac{dy}{dx}=-{\left(\frac{y}{x}\right)}^{\frac{1}{3}}$

$=-\frac{\sqrt{a^{\frac{2}{3}}-x^{\frac{2}{3}}}}{x^{\frac{1}{3}}}$