If f(x)= csc 3 x  and g(x) = sqrt(2x-3 , how do you differentiate f(g(x))  using the chain rule?

$f ' \left(g \left(x\right)\right) = - \frac{3}{\sqrt{2 x - 3}} \cdot \csc \left(3 \sqrt{2 x - 3}\right) \cdot \cot \left(3 \sqrt{2 x - 3}\right)$

Explanation:

Given $f \left(x\right) = \csc 3 x$ and $g \left(x\right) = \sqrt{2 x - 3}$

$f \left(g \left(x\right)\right) = \csc 3 g \left(x\right)$
$f \left(g \left(x\right)\right) = \csc 3 \sqrt{2 x - 3}$

The formula for derivative of $\csc u :$

$\frac{d}{\mathrm{dx}} \left(\csc u\right) = - \csc u \cdot \cot u \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Let $u = 3 \sqrt{2 x - 3}$

Take note:
$\frac{d}{\mathrm{dx}} \left(3 \sqrt{2 x - 3}\right) = 3 \cdot \frac{1}{2 \sqrt{2 x - 3}} \cdot 2$

$f ' \left(g \left(x\right)\right) =$
$- \csc \left(3 \sqrt{2 x - 3}\right) \cdot \cot \left(3 \sqrt{2 x - 3}\right) \cdot 3 \cdot \frac{1}{2 \sqrt{2 x - 3}} \cdot 2$

$f ' \left(g \left(x\right)\right) = - \frac{3}{\sqrt{2 x - 3}} \cdot \csc \left(3 \sqrt{2 x - 3}\right) \cdot \cot \left(3 \sqrt{2 x - 3}\right)$