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    Choose the correct answer out of 4 options given against each Question

If y=\frac{\sin x+\cos x}{\sin x-\cos x}   then  \frac{dy}{dx}_{at\: \: x=0} is  

 

A. –2
B. 0
C. \frac{1}{2}
D. does not exist

Answers (1)

\\ y=\frac{\sin x+\cos x}{\sin x-\cos x} \\ \\ \frac{dy}{dx}=\frac{ \left( \sin x-\cos x \right) \left( \cos x-\sin x \right) - \left( \sin x+\cos x \right) \left( \sin x+\cos x \right) }{ \left( \sin x-\cos x \right) ^{2}} \\ \\ =\frac{- \left( \sin ^{2}x+\cos ^{2}x-2\sin x\cos x \right) - \left( \sin ^{2}x+\cos ^{2}x+2\sin x\cos x \right) }{ \left( \sin x-\cos x \right) ^{2}}=-\frac{2}{ \left( \sin x-\cos x \right) ^{2}} \\ \\ \left( \frac{dy}{dx} \right) _{x=0}=-\frac{2}{ \left( -1 \right) ^{2}}=-2 \\ \\

Hence, the answer is option A

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