Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Differentiate the functions w.r.t. x. cos x cos 2x cos 3x

 Q1   Differentiate the functions w.r.t. x.  \cos x . \cos 2x .\cos 3x

 

Answers (1)

best_answer

Given function is
y=\cos x . \cos 2x .\cos 3x
Now, take log on both sides
\log y=\log (\cos x . \cos 2x .\cos 3x)\\ \log y = \log \cos x + \log \cos 2x + \log \cos 3x
Now, differentiation w.r.t. x
\log y=\log (\cos x . \cos 2x .\cos 3x)\\ \frac{d(\log y )}{dx} = \frac{\log \cos x}{dx} + \frac{\log \cos 2x}{dx} + \frac{\log \cos 3x}{dx}\\ \frac{1}{y}.\frac{dy}{dx} = (-\sin x)\frac{1}{\cos x}+(-2\sin 2x)\frac{1}{\cos 2x}+(-3\sin3x).\frac{1}{\cos3x}\\ \frac{1}{y}\frac{dy}{dx} = -(\tan x+\tan 2x+\tan 3x) \ \ \ \ \ \ (\because \frac{\sin x }{\cos x} =\tan x)\\ \frac{dy}{dx}=-y(\tan x+\tan 2x+\tan 3x)\\ \frac{dy}{dx}= -\cos x\cos 2x\cos 3x(\tan x+\tan 2x+\tan 3x)
There, the answer is  -\cos x\cos 2x\cos 3x(\tan x+\tan 2x+\tan 3x)

Posted by

Gautam harsolia

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Latest Question