Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

3 X X + Tan X - Mathematics

$\frac{3^x}{x + \tan x}$

Solution

$\text{ Let } u = 3^x ; v = x + \tan x$
$\text{ Then }, u' = 3^x \log 3; v' = 1 + \sec^2 x$
$\text{ By quotient rule, we have }:$
$\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}$
$\frac{d}{dx}\left( \frac{3^x}{x + \tan x} \right) = \frac{\left( x + \tan x \right) 3^x \log 3 - 3^x \left( 1 + \sec^2 x \right)}{\left( x + \tan x \right)^2}$
$= \frac{3^x \left[ \left( x + \tan x \right) \log 3 - \left( 1 + \sec^2 x \right) \right]}{\left( x + \tan x \right)^2}$

Concept: The Concept of Derivative - Algebra of Derivative of Functions
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Exercise 30.5 | Q 19 | Page 44