Question

If $y=\frac{sinx+cosx}{sinx-cosx}$, then $\frac{dy}{dx}$ at $x=0$ is

• A. -2
• B. 1/2
• C. does not exist

Answer 1

The correct option is A -2

$y=\frac{sinx+cosx}{sinx-cosx}$

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=\frac{(sinx-cosx)\times \frac{d}{dx}(sinx+cosx)-(sinx+cosx)\times \frac{d}{dx}(sinx-cosx)}{(sinx-cosx{)}^2}$

$=\frac{(sinx-cosx)\times \left[\frac{d}{dx}\right(sinx)+\frac{d}{dx}(cosx\left)\right]-(sinx+cosx)\times \left[\frac{d}{dx}\right(sinx)-\frac{d}{dx}(cosx\left)\right]}{(sinx-cosx{)}^2}$

$=\frac{(sinx-cosx)(cosx-sinx)-(sinx+cosx)\left(2sinxcosx\right)}{(sinx-cosx{)}^2}$

$=\frac{-(co{s}^{2}x+si{n}^{2}x-2cosxsinx)-(si{n}^{2}x+co{s}^{2}x+2sinxcosx)}{(sinx-cosx{)}^2}$
$=\frac{-1+2cosxsinx-1-2sinxcosx}{(sinx-cosx{)}^2}=\frac{-2}{(sinx-cosx{)}^2}$

Putting $x=0,$ we get

${\left(\frac{dy}{dx}\right)}_{x=0}=\frac{-2}{(sin0-cos0{)}^2}=\frac{-2}{(0-1{)}^2}=-2$

Thus, $\frac{dy}{dx}$ at $x=0$ is -2

Hence, the correct answer is option (a).