# What is the derivative of csc(3x)?

Oct 27, 2017

I'll add some detail: how you get there.

#### Explanation:

...you use the rule for finding the derivative of the quotient of 2 functions.

If $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)} , t h e n f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - g \left(x\right) h ' \left(x\right)}{h {\left(x\right)}^{2}}$

So, you start by rewriting $\csc \left(3 x\right) = \frac{1}{\sin} \left(3 x\right)$

Here, g(x) = 1, g'(x) = 0, h(x) = sin(3x), h'(x) = 3cos(3x)

Put all this together:

$\frac{d}{\mathrm{dx}} \left(\frac{1}{\sin} \left(3 x\right)\right) = \frac{- 3 \cos \left(3 x\right)}{{\sin}^{2} \left(3 x\right)}$

$= \frac{- 3 \cos \left(3 x\right)}{\sin \left(3 x\right) \sin \left(3 x\right)}$

$= - 3 \cot \left(3 x\right) \csc \left(3 x\right)$

GOOD LUCK