# How do you find the derivative of ln(tanx)?

Use the chain rule and use $\frac{d}{\mathrm{dx}} \left(\ln u\right) = \frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}}$.
We'll also need $\frac{d}{\mathrm{dx}} \left(\tan x\right) = {\sec}^{2} x$
$\frac{d}{\mathrm{dx}} \left(\ln \left(\tan x\right)\right) = \frac{1}{\tan} x \frac{d}{\mathrm{dx}} \left(\tan x\right) = \frac{1}{\tan} x {\sec}^{2} x$
$\frac{d}{\mathrm{dx}} \left(\ln \left(\tan x\right)\right) = \frac{1}{\sin \frac{x}{\cos} x} \frac{1}{{\cos}^{2} x} = \frac{1}{\sin} x \frac{1}{\cos} x = \csc x \sec x$