question_answer

If \[{{x}^{2/3}}+{{y}^{2/3}}={{a}^{2/3}},\] then find \[\frac{dy}{dx}.\]

Answer:

We have, \[{{x}^{2/3}}+{{y}^{2/3}}={{a}^{2/3}}\]             On differentiating both sides w.r.t. x, we get                         \[\frac{2}{3}{{x}^{2/3-1}}+\frac{2}{3}{{y}^{2/3-1}}\frac{dy}{dx}=0\]             \[\Rightarrow \]   \[{{y}^{-1/3}}\frac{dy}{dx}=-{{x}^{^{-1/3}}}\]             \[\Rightarrow \]   \[\frac{dy}{dx}=\frac{-{{x}^{^{-1/3}}}}{{{y}^{^{-1/3}}}}=-\,{{\left( \frac{y}{x} \right)}^{1/3}}\]