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\( u=\frac{1}{1+x^{2}}, \quad v=\tan ^{-1} x \). |
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Answer» u \(=\frac1{1+x^2}\) ∴ \(\frac{du}{dx}\) \(=\frac{-1}{(1+x^2)^2}\) d/dx(1+x2) (By chain rule and d/dx (1/x) = -1/x2) \(=\frac{-2x}{(1+x^2)^2}\) (∵ d/dx (1+x2) = d/dx 1 + d/dx x2 = 2x) And V = tan-1x ∵ \(\frac{dV}{dx}=\frac1{1+x^2}\) Now, \(\frac{du}{dV}=\cfrac{\frac{du}{dx}}{\frac{dV}{dx}}=\cfrac{\frac{-2x}{(1+x^2)^2}}{\frac{1}{1+x^2}}=\frac{-2x}{1+x^2}\) |
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