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If \( y=\sin ^{-1}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right) \) find dy/ dx. |
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Answer» y = sin-1\((\frac{x^2-y^2}{x^2+y^2})\) Differentiate both sides w.r.t. x we get \(\frac{dy}{dx}\) = \(\cfrac{1}{\sqrt{1-(\frac{x^2-y^2}{x^2+y^2})^2}}\) \(\times\cfrac{(x^2+y^2)(2x-2y\frac{dy}{dx})(x^2-y^2)(2x+2y\frac{dy}{dx})}{(x^2+y^2)^2}\) \(=\frac{x^2+y^2}{\sqrt{(x^2+y^2)^2-(x^2-y^2)^2}}\)\(\times\cfrac{(2x^3+2xy^2-2x^3+2xy^2)-2y\frac{dy}{dx}(x^2+y^2+x^2-y^2)}{(x^2+y^2)^2}\) \(=\frac1{\sqrt{4x^2y^2}}\times\cfrac{4xy^2-4x^2y\frac{dy}{dx}}{x^2+y^2}\) \(=\frac1{2xy}\times\cfrac{2xy(2y-2x\frac{dy}{dx})}{x^2+y^2}\) ⇒ \(\frac{dy}{dx}=\cfrac{2y-2x\frac{dy}{dx}}{x^2+y^2}\) ⇒ \(\frac{dy}{dx}+\frac{2x}{x^2+y^2}\frac{dy}{dx}=\frac{2y}{x^2+y^2}\) ⇒ \(\frac{dy}{dx}(1 + \frac{2x}{x^2+y^2})=\frac{2y}{x^2+y^2}\) ⇒ \(\frac{dy}{dx}=\frac{2y}{x^2+y^2+2x}\) |
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