What is the derivative of f(x) = 1- sec x/ tan x?

Apr 5, 2018

${f}^{'} \left(x\right) = \csc x \cot x$

Explanation:

Here,

$f \left(x\right) = 1 - \sec \frac{x}{\tan} x$

$\text{Applying "color(blue)" Quotient Rule}$

${f}^{'} \left(x\right) = 0 - \frac{\tan x \frac{d}{\mathrm{dx}} \left(\sec x\right) - \sec x \frac{d}{\mathrm{dx}} \left(\tan x\right)}{\tan x} ^ 2$

$= - \frac{\tan x \sec x \tan x - \sec x {\sec}^{2} x}{\tan} ^ 2 x$

$= - \sec x \left[\frac{{\tan}^{2} x - {\sec}^{2} x}{\tan} ^ 2 x\right]$

$= \sec x \left[\frac{{\sec}^{2} x - {\tan}^{2} x}{\tan} ^ 2 x\right]$,where, ${\sec}^{2} x - {\tan}^{2} x = 1$

$= \frac{\sec x}{\tan} ^ 2 x$

$= \frac{\frac{1}{\cos} x}{{\sin}^{2} \frac{x}{\cos} ^ 2 x}$

$= \cos \frac{x}{\sin} ^ 2 x$

$= \frac{1}{\sin} x \cdot \cos \frac{x}{\sin} x$

$= \csc x \cot x$

Apr 5, 2018

${f}^{'} \left(x\right) = \csc x \cot x$

Explanation:

Here,

$f \left(x\right) = 1 - \sec \frac{x}{\tan} x$

$= 1 - \frac{\frac{1}{\cos} x}{\sin \frac{x}{\cos} x}$

$= 1 - \frac{1}{\sin} x$

$= 1 - \csc x$

${f}^{'} \left(x\right) = 0 - \left(- \csc x \cot x\right)$

$= \csc x \cot x$