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