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Question
If \[\sin^{- 1} \left( \frac{x^2 - y^2}{x^2 + y^2} \right) = \text { log a then } \frac{dy}{dx}\] is equal to _____________ .
Options
\[\frac{x^2 - y^2}{x^2 + y^2}\]
`y/x`
`x/y`
none of these
Solution
\[\frac{y}{x}\]
\[\text { We have }, \sin^{- 1} \left( \frac{x^2 - y^2}{x^2 + y^2} \right) = \log a\]
\[ \Rightarrow \frac{x^2 - y^2}{x^2 + y^2} = \sin \log a\]
\[\Rightarrow \frac{\left( x^2 + y^2 \right)\left( 2x - 2y\frac{dy}{dx} \right) - \left( x^2 - y^2 \right)\left( 2x + 2y\frac{dy}{dx} \right)}{\left( x^2 + y^2 \right)^2} = 0\]
\[ \Rightarrow \frac{2 x^3 - 2 x^2 y\frac{dy}{dx} + 2x y^2 - 2 y^3 \frac{dy}{dx} - 2 x^3 - 2 x^2 y\frac{dy}{dx} + 2x y^2 + 2 y^3 \frac{dy}{dx}}{\left( x^2 + y^2 \right)^2} = 0\]
\[ \Rightarrow - 4 x^2 y\frac{dy}{dx} + 4x y^2 = 0\]
\[ \Rightarrow - 4 x^2 y\frac{dy}{dx} = - 4x y^2 \]
\[ \Rightarrow \frac{dy}{dx} = \frac{4x y^2}{4 x^2 y}\]
\[ \therefore \frac{dy}{dx} = \frac{y}{x}\]
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