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If Sec ( X + Y X − Y ) = a Prove that D Y D X = Y X ? - Mathematics

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Question

If \[\sec \left( \frac{x + y}{x - y} \right) = a\] Prove that  \[\frac{dy}{dx} = \frac{y}{x}\] ?

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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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Chapter 11: Differentiation - Exercise 11.04 [Page 75]

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RD Sharma Mathematics [English] Class 12
Chapter 11 Differentiation
Exercise 11.04 | Q 18 | Page 75

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