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Find:d/dx cot-1((1 - x)/(1 + x))\(\frac{d}{dx}cot^{-1}(\frac{1-x}{1+x})\) |
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Answer» \(\frac{d}{dx}cot^{-1}(\frac{1-x}{1+x})\) = \(\cfrac{-1}{1+(\frac{1-x}{1+x})^2}\) \(\frac{d}{dx}(\frac{1-x}{1+x})\) = \(\frac{-(1+x)^2}{(1+x)^2+(1-x)^2}\)\(\times\) \(\frac{(1+x)\times-1-(1-x)\times1}{(1+x)^2}\) = \(\frac{-(1+x)^2}{(1+x)^2+(1-x)^2}\) \(\times\)\(\frac{-1-x-1+x}{(1+x)^2}\) = \(\frac1{1+x^2}\) ∴ \(\frac{d}{dx}\)cot -1(\(\frac{1-x}{1+x}\)) = \(\frac1{1+x^2}\) |
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