# Differentiate cos(x-2)/sinx?

May 10, 2018

$f \left(x\right) = \cos \frac{x - 2}{\sin} x = \frac{\cos x \cos 2 - \sin x \sin 2}{\sin} x$
$f \left(x\right) = \cos 2 \cot x - \sin 2$
$\therefore f ' \left(x\right) = \cos 2 \left(- {\csc}^{2} x\right) - 0 = - \cos 2 {\csc}^{2} x = - \cos \frac{2}{\sin} ^ 2 x$
OR

#### Explanation:

We take,

$f \left(x\right) = \frac{\cos \left(x - 2\right)}{\sin} x$

Diff.w.r.t. $x$,$\text{using "color(blue)"Quotient Rule :}$

color(blue)(d/(dx)(u*v)=(v*(du)/(dx)-u*(dv)/(dx))/v^2

$f ' \left(x\right) = \frac{\sin x \cdot \frac{d}{\mathrm{dx}} \left(\cos \left(x - 2\right)\right) - \cos \left(x - 2\right) \frac{d}{\mathrm{dx}} \left(\sin x\right)}{{\left(\sin x\right)}^{2}}$

$f ' \left(x\right) = \frac{\sin x \left(- \sin \left(x - 2\right)\right) - \cos \left(x - 2\right) \cos x}{\sin} ^ 2 x$

$f ' \left(x\right) = - \frac{\left[\cos \left(x - 2\right) \cos x + \sin \left(x - 2\right) \sin x\right]}{\sin} ^ 2 x$

$f ' \left(x\right) = - \frac{\cos \left(\left(x - 2\right) - x\right)}{\sin} ^ 2 x = - \cos \frac{- 2}{\sin} ^ 2 x$

$f ' \left(x\right) = - \cos \frac{2}{\sin} ^ 2 x$