# What is the derivative of y=ln(sec(x)+tan(x))?

Jul 26, 2014

Answer: $y ' = \sec \left(x\right)$

Full explanation:

Suppose, $y = \ln \left(f \left(x\right)\right)$

Using chain rule, $y ' = \frac{1}{f} \left(x\right) \cdot f ' \left(x\right)$

Similarly, if we follow for the problem, then

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

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

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

$y ' = \sec \left(x\right)$

Apr 18, 2015

Will give you a personal video explanation of how it's done...

Alternatively, you can use these workings...

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

${e}^{y} = \sec x + \tan x$

${e}^{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \sec x \tan x + {\sec}^{2} x$

${e}^{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \sec x \left(\sec x + \tan x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sec x \left(\sec x + \tan x\right)}{e} ^ y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sec x \left(\sec x + \tan x\right)}{\left(\sec x + \tan x\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sec x$