Question

If $y=\frac{\sin x+\cos x}{\sin x-\cos x}$ find $\frac{dy}{dx}$ at $x=\frac{\pi }{4}$

Answer 1

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

$\frac{\frac{sinx.1cosx}{cosx}}{\frac{sinx-cosx}{cosx}}$

$y=\frac{tanx+1}{tanx-1}$

$y=\frac{-(1+tanx)}{1-tanx}$

$\Rightarrow y=-tan(\frac{\pi }{4}+x)$

$\frac{dy}{dx}=-se{c}^2(\frac{\pi }{4}+x)$

$\frac{dy}{dx}{]}_{a+x}=\frac{\pi }{4}$ $=-se{c}^2\frac{\pi }{2}=-\infty$