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Question
If \[\sec \left( \frac{x + y}{x - y} \right) = a\] Prove that \[\frac{dy}{dx} = \frac{y}{x}\] ?
Solution
\[\text{ We have }, \sec\left( \frac{x + y}{x - y} \right) = a\]
\[ \Rightarrow \frac{x + y}{x - y} = \sec^{- 1} \left( a \right)\]
Differentiate with respect to x, we get,
\[\Rightarrow \left[ \frac{\left( x - y \right)\frac{d}{dx}\left( x + y \right) - \left( x + y \right)\frac{d}{dx}\left( x - y \right)}{\left( x - y \right)^2} \right] = 0\]
\[ \Rightarrow \left( x - y \right) \left( 1 + \frac{d y}{d x} \right) - \left( x + y \right) \left( 1 - \frac{d y}{d x} \right) = 0\]
\[ \Rightarrow \left( x - y \right) + \left( x - y \right)\frac{d y}{d x} - \left( x + y \right) + \left( x + y \right)\frac{d y}{d x} = 0\]
\[ \Rightarrow \frac{d y}{d x}\left[ x - y + x + y \right] = x + y - x + y\]
\[ \Rightarrow \frac{d y}{d x}\left( 2x \right) = 2y\]
\[ \Rightarrow \frac{d y}{d x} = \frac{y}{x}\]
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