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Question
Find \[\frac{dy}{dx}\] in the following case \[x^{2/3} + y^{2/3} = a^{2/3}\] ?
Solution
\[\text{We have }, x^\frac{2}{3} + y^\frac{2}{3} = a^\frac{2}{3} \]
Differentiating it with respect to x, we get,
\[\frac{d}{dx}\left( x^\frac{2}{3} \right) + \frac{d}{dx}\left( y^\frac{2}{3} \right) = \frac{d}{dx}\left( a^\frac{2}{3} \right)\]
\[ \Rightarrow \frac{2}{3} \left( x \right)^{\frac{2}{3} - 1} + \frac{2}{3} \left( y \right)^{\frac{2}{3} - 1} \frac{d y}{d x} = 0\]
\[ \Rightarrow \frac{2}{3} \left( x \right)^\frac{- 1}{3} + \frac{2}{3} \left( y \right)^\frac{- 1}{3} \frac{d y}{d x} = 0\]
\[ \Rightarrow \frac{2}{3} \left( y \right)^\frac{- 1}{3} \frac{d y}{d x} = - \frac{2}{3} \left( x \right)^\frac{- 1}{3} \]
\[ \Rightarrow \frac{d y}{d x} = - \frac{2}{3} \left( x \right)^\frac{- 1}{3} \times \frac{3}{2 y^\frac{- 1}{3}}\]
\[ \Rightarrow \frac{d y}{d x} = - \frac{x^\frac{- 1}{3}}{y^\frac{- 1}{3}}\]
\[ \Rightarrow \frac{d y}{d x} = - \frac{y^\frac{1}{3}}{x^\frac{1}{3}}\]
\[ \Rightarrow \frac{d y}{d x} = - \left( \frac{y}{x} \right)^\frac{1}{3} \]
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