# What is the derivative of (12/sinx) + (1/cotx)?

Sep 16, 2015

−12cosecx⋅cotx+sec^2x

#### Explanation:

1/$\sin x$ = $\cos e c x$ similarly $\frac{1}{\cot} x$ = $\tan x$
the question changes to $12 \cos e c x + \tan x$
the derivative is $- 12 \cos e c x \cdot \cot x + {\sec}^{2} x$ by (u/v rule of differentiation)
$\frac{d}{\mathrm{dx}}$ $\left(\frac{1}{\sin} x\right)$ = $\frac{1}{\sin} ^ 2 x$ $\left(- 1 \cdot \cos x\right)$
= $- \cos \frac{x}{\sin} x$ $\frac{1}{\sin} x$
=$- \cos e c x \cot x$
$\frac{d}{\mathrm{dx}}$ $\tan x$= $\frac{d}{\mathrm{dx}} \left(\sin \frac{x}{\cos} x\right)$
=$\frac{1}{\cos} ^ 2 x$ $\left(\cos x \cdot \cos x - \sin x \cdot \left(- \sin x\right)\right)$
=$\frac{{\cos}^{2} x + {\sin}^{2} x}{\cos} ^ 2 x$
=$\frac{1}{\cos} ^ 2 x$=${\sec}^{2} x$