# How do you differentiate y=ln(secx tanx)?

Mar 15, 2015

Use the derivative of $\ln x$ and the Chain Rule.

$\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}$, so (by the chain rule)

$\frac{d}{\mathrm{dx}} \left(\ln \left(u\right)\right) = \frac{1}{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$y = \ln \left(\sec x \tan x\right)$

$y ' = \frac{1}{\sec x \tan x} \cdot \frac{d}{\mathrm{dx}} \left(\sec x \tan x\right)$

$= \frac{1}{\sec x \tan x} \cdot \left(\left(\sec x \tan x\right) \tan x + \sec x \cdot {\sec}^{2} x\right)$

(I use the product rule in the form: $\frac{d}{\mathrm{dx}} \left(F S\right) = F ' S + F S '$)

$y ' = \frac{1}{\sec x \tan x} \left(\sec x {\tan}^{2} x + {\sec}^{3} x\right)$

There are many other ways to write $\frac{1}{\sec x \tan x} \left(\sec x {\tan}^{2} x + {\sec}^{3} x\right)$. One of the more obvious is:

$\frac{\sec x {\tan}^{2} x + {\sec}^{3} x}{\sec x \tan x} = \frac{{\tan}^{2} x + {\sec}^{2} x}{\tan} x$ which

could also be written: $\tan x + {\sec}^{2} \frac{x}{\tan} x = \tan x + \sec x \csc x$. or, factoring out $\frac{1}{\cos} x$, we could write $\sec x \left(\sin x + \csc x\right)$ and so on.
(And trig students ask why we have to do identities).