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Find the derivative of \( f ( x )=\frac{x}{1+\tan x} \). |
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Answer» f(x) = \(\frac{x}{1+tan x}\) \(\therefore\) f'(x) = \(\cfrac{(1+tan x)\frac{dx}{dx}-x\frac{d}{dx}(1+tanx)}{(1+tanx)^2}\) \(\Big(\because\) \(\frac{d}{dx}\frac{u}{v}=\cfrac{v\frac{dy}{dx}-u\frac{dv}{dx}}{v^2}\)\(\Big)\) \(=\frac{1+tan x-x.sec^2x}{(1+tan x)^2}\) |
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