# What is the derivative of this function f(x)=1/sinx?

May 28, 2018

The answer is $= - \cos \frac{x}{\sin} ^ 2 x$

#### Explanation:

The function is

$f \left(x\right) = \frac{1}{\sin} x = {\left(\sin x\right)}^{-} 1$

The derivative is

$f ' \left(x\right) = \left({\left(\sin x\right)}^{-} 1\right) ' = \left(- {\left(\sin x\right)}^{-} 2\right) \cdot \cos x$

$= - \cos \frac{x}{\sin} ^ 2 x$

May 28, 2018

$- \cot x \csc x$.

#### Explanation:

We will use the Quotient Rule (QR) to find $f ' \left(x\right)$.

QR : $f \left(x\right) = g \frac{x}{h \left(x\right)} , \text{ then, } f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2$.

$\therefore f \left(x\right) = \frac{1}{\sin} x \Rightarrow f ' \left(x\right) = \frac{\sin x \cdot \frac{d}{\mathrm{dx}} \left(1\right) - 1 \cdot \frac{d}{\mathrm{dx}} \left(\sin x\right)}{\sin x} ^ 2$,

$= \frac{\sin x \cdot 0 - 1 \cdot \cos x}{\sin} ^ 2 x$,

$= - \cos \frac{x}{\sin} ^ 2 x$,

$= - \cos \frac{x}{\sin} x \cdot \frac{1}{\sin} x$,

$= - \cot x \csc x$.