# X Tan X Sec X + Tan X - Mathematics

$\frac{x \tan x}{\sec x + \tan x}$

#### Solution

$\text{ Let } u = x \tan x; v = \sec x + \tan x$
$\text{ Then }, u' = x \sec^2 x + \tan x; v' = \sec x \tan x + \sec^2 x$
$\text{ Using the quotient rule }:$
$\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}$
$\frac{d}{dx}\left( \frac{x\tan x}{\sec x + \tan x} \right) = \frac{\left( \sec x + \tan x \right)\left( x \sec^2 x + \tan x \right) - x \tan x\left( \sec x \tan x + \sec^2 x \right)}{\left( \sec x + \tan x \right)^2}$
$= \frac{x \sec^3 x + x \sec^2 x\tan x + \sec x \tan x + \tan^2 x - x \sec x \tan^2 x - x \tan x \sec^2 x}{\left( \sec x + \tan x \right)^2}$
$= \frac{\left( \sec x + \tan x \right)\left( x \sec^2 x + \tan x \right) - x \tan x \sec x\left( \sec x + \tan x \right)}{\left( \sec x + \tan x \right)^2}$
$= \frac{x \sec^2 x + \tan x - x \tan x \sec x}{\sec x + \tan x}$
$= \frac{x \sec x\left( \sec x - \tan x \right) + \tan x}{\sec x + \tan x}$

Concept: The Concept of Derivative - Algebra of Derivative of Functions
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Exercise 30.5 | Q 10 | Page 44