Question

Mark the correct alternative in each of the following:

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

$y=\frac{\sin x+\cos x}{\sin x-\cos x}$

Differentiating both sides with respect to x, we get

$$\begin{gathered} \frac{dy}{dx}=\frac{\left(\sin x-\cos x\right)\times {\displaystyle \frac{d}{dx}}\left(\sin x+\cos x\right)-\left(\sin x+\cos x\right)\times {\displaystyle \frac{d}{dx}}\left(\sin x-\cos x\right)}{{\left(\sin x-\cos x\right)}^2}\left(\mathrm{Quotient}\mathrm{rule}\right) \\ =\frac{\left(\sin x-\cos x\right)\times \left[{\displaystyle \frac{d}{dx}}\left(\sin x\right)+{\displaystyle \frac{d}{dx}}\left(\cos x\right)\right]-\left(\sin x+\cos x\right)\times \left[{\displaystyle \frac{d}{dx}}\left(\sin x\right)-{\displaystyle \frac{d}{dx}}\left(\cos x\right)\right]}{{\left(\sin x-\cos x\right)}^2} \\ =\frac{\left(\sin x-\cos x\right)\left(\cos x-\sin x\right)-\left(\sin x+\cos x\right)\left(\cos x+\sin x\right)}{{\left(\sin x-\cos x\right)}^2} \\ =\frac{-\left({\cos }^{2}x+{\sin }^{2}x-2\cos x\sin x\right)-\left({\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x\right)}{{\left(\sin x-\cos x\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{-1+2\cos x\sin x-1-2\sin x\cos x}{{\left(\sin x-\cos x\right)}^2} \\ =\frac{-2}{{\left(\sin x-\cos x\right)}^2} \end{gathered}$$

Putting x = 0, we get

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

Thus, $\frac{dy}{dx}$ at x = 0 is −2.

Hence, the correct answer is option (a).