Question

If $y=\left(\frac{2-3cosx}{sinx}\right),$ find $\frac{dy}{dx}$ at $x=\frac{\pi }{4}$

Answer 1

We have,

$y=\frac{2-3cosx}{sinx}$

$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{2-3cosx}{sinx}\right)$

$=\frac{d}{dx}(2cosecx=3cosx)$

$=2\frac{d}{dx}\left(cosecx\right)-33\frac{d}{dx}\left(cotx\right)$

$=-2cosecx.cotx+3cose{c}^{2}x\frac{dy}{dx}$ at $x=\frac{\pi }{4}$

$=-2cosec\frac{\pi }{4}.cot\frac{\pi }{4}+3cose{c}^2\frac{\pi }{4}$

$=-2\sqrt{2}-1+3.2$

$=-2\sqrt{2}+6$

$=6-2\sqrt{2}$