Question

If $y=\frac{sinx+cosx}{sinx-cosx}$, 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, $y=\frac{sinx+cosx}{sinx-cosx}$
$\left\{\left(sinx-cosx\right)\frac{d}{dx}\left(sinx+cosx\right)\right\}$
$\Rightarrow \frac{dy}{dx}=\frac{-\left(sinx+cosx\right)\frac{d}{dx}\left(sinx-cosx\right)}{{\left(sinx-cosx\right)}^2}$
$=\frac{\left(sinx-cosx\right)\left(cosx-sinx\right)-\left(sinx+cosx\right)\left(cosx+sinx\right)}{{\left(sinx-cosx\right)}^2}$
$=\frac{-si{n}^{2}x-co{s}^{2}x+2sinxcosx-si{n}^{2}x-co{s}^{2}x-2sinxcosx}{{\left(sinx-cosx\right)}^2}$
$=\frac{-2}{{\left(sinx-cosx\right)}^2}$
$\Rightarrow \frac{dy}{dx}{|}_{atx=0}$ $=\frac{-2}{{\left(sin0-cos0\right)}^2}=\frac{-2}{{\left(0-1\right)}^2}=-2$