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

Answer 1

Answer: (A)

We have, $y=\frac{sin\,x+cos\,x}{sin\,x-cos\,x}$
$\left\{\left(sin\,x-cos\,x\right) \frac{d}{dx}\left(sin\,x+cos\,x\right)\right\}$
$\Rightarrow \frac{dy}{dx}=\frac{-\left(sin\,x+cos\,x\right) \frac{d}{dx}\left(sin\,x-cos\,x\right)}{\left(sin\,x-cos\,x\right)^{2}}$
$=\frac{\left(sin\, x - cos\, x \right)\left(cos \,x - sin\, x\right) - \left(sin\, x+ cos \,x\right)\left(cos \,x +sin\, x\right)}{\left(sin\,x-cos\,x\right)^{2}}$
$=\frac{- sin^{2}\, x - cos^{2}\,x + 2\, sin\, x \,cos\, x - sin^{2}\, x - cos^{2}\, x -2\, sin \,x \,cos \,x}{\left(sin\,x-cos\,x\right)^{2}}$
$=\frac{-2}{\left(sin\,x-cos\,x\right)^{2}}$
$\Rightarrow \frac{dy}{dx}\bigg|_{at\,x=0}$ $=\frac{-2}{\left(sin\,0-cos\,0\right)^{2}}=\frac{-2}{\left(0-1\right)^{2}}=-2$