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Find \( \frac{d y}{d x} \), if \( y=\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right) \). |
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Answer» y = tan-1(\(\frac{cos x - sin x}{cos x + sin x}\)) = tan-1(\(\frac{1-tan x}{1+tan x}\)) (On dividing numerator and denominator by cos x) = tan-1(\(\frac{tan \pi/4-tan x}{1+tan \pi/4 tan x}\)) = tan-1(tan(\(\pi/4-x\))) = \(\pi/4-x\) \(\therefore\) \(\frac{dy}{dx}=\frac{d}{dx}(\frac{\pi}4-x)=-1\) |
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