Question

Find $\frac{dy}{dx}$ in each of the following cases:

$x^{2/3}+y^{2/3}=a^{2/3}$

Answer 1

$\mathrm{We}\mathrm{have},x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$
Differentiating it with respect to x, we get,

$$\begin{gathered} \frac{d}{dx}\left(x^{\frac{2}{3}}\right)+\frac{d}{dx}\left(y^{\frac{2}{3}}\right)=\frac{d}{dx}\left(a^{\frac{2}{3}}\right) \\ \Rightarrow \frac{2}{3}{\left(x\right)}^{\frac{2}{3}-1}+\frac{2}{3}{\left(y\right)}^{\frac{2}{3}-1}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=0 \\ \Rightarrow \frac{2}{3}{\left(x\right)}^{\frac{-1}{3}}+\frac{2}{3}{\left(y\right)}^{\frac{-1}{3}}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=0 \\ \Rightarrow \frac{2}{3}{\left(y\right)}^{\frac{-1}{3}}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=-\frac{2}{3}{\left(x\right)}^{\frac{-1}{3}} \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=-\frac{2}{3}{\left(x\right)}^{\frac{-1}{3}}\times \frac{3}{2y^{{\displaystyle \frac{-1}{3}}}} \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=-\frac{x^{{\displaystyle \frac{-1}{3}}}}{y^{{\displaystyle \frac{-1}{3}}}} \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=-\frac{y^{{\displaystyle \frac{1}{3}}}}{x^{{\displaystyle \frac{1}{3}}}} \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=-{\left(\frac{y}{x}\right)}^{\frac{1}{3}} \end{gathered}$$