3 X X + Tan X - Mathematics

Question

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

Solution

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

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The Concept of Derivative - Algebra of Derivative of Functions

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Q 19 Q 18 Q 20

Chapter 30: Derivatives - Exercise 30.5 [Page 44]

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RD Sharma Mathematics [English] Class 11

Chapter 30 Derivatives
Exercise 30.5 | Q 19 | Page 44

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3 X X + Tan X - Mathematics

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