Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

If $y = \log \sqrt {\dfrac{{1 - \cos 3x}}{{1 + \cos 3x}}} $, find $\dfrac{{dy}}{{dx}}.$

seo-qna
Last updated date: 23rd May 2024
Total views: 436.5k
Views today: 12.36k
Answer
VerifiedVerified
436.5k+ views
Hint: Convert trigonometric terms in their half angles.

As we know the differentiation of $\log \left( {ax + b} \right) = \dfrac{1}{{ax + b}}\left( {\dfrac{d}{{dx}}\left( {ax + b} \right)} \right)$, so, use this formula the differentiation of given equation is
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {\dfrac{{1 - \cos 3x}}{{1 + \cos 3x}}} }}\left( {\dfrac{d}{{dx}}\left( {\sqrt {\dfrac{{1 - \cos 3x}}{{1 + \cos 3x}}} } \right)} \right)..............\left( 1 \right)$
Now we know$\left( {1 - \cos ax} \right) = 2{\sin ^2}\left( {\dfrac{{ax}}{2}} \right),{\text{ }}\left( {1 + \cos ax} \right) = 2{\cos ^2}\left( {\dfrac{{ax}}{2}} \right)$, so, use this property equation 1 becomes
$
  \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {\dfrac{{2{{\sin }^2}\left( {\dfrac{{3x}}{2}} \right)}}{{2{{\cos }^2}\left( {\dfrac{{3x}}{2}} \right)}}} }}\left( {\dfrac{d}{{dx}}\left( {\sqrt {\dfrac{{2{{\sin }^2}\left( {\dfrac{{3x}}{2}} \right)}}{{2{{\cos }^2}\left( {\dfrac{{3x}}{2}} \right)}}} } \right)} \right) \\
  \dfrac{{\sin x}}{{\cos x}} = \tan x \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {{{\tan }^2}\left( {\dfrac{{3x}}{2}} \right)} }}\left( {\dfrac{d}{{dx}}\left( {\sqrt {{{\tan }^2}\left( {\dfrac{{3x}}{2}} \right)} } \right)} \right) \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\tan \left( {\dfrac{{3x}}{2}} \right)}}\dfrac{d}{{dx}}\left( {\tan \left( {\dfrac{{3x}}{2}} \right)} \right) \\
$
Now we know $\tan x$differentiation is ${\sec ^2}x$
\[
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\tan \left( {\dfrac{{3x}}{2}} \right)}}{\sec ^2}\left( {\dfrac{{3x}}{2}} \right)\left( {\dfrac{d}{{dx}}\dfrac{{3x}}{2}} \right) \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\dfrac{{\sin \left( {\dfrac{{3x}}{2}} \right)}}{{\cos \left( {\dfrac{{3x}}{2}} \right)}}}}\dfrac{1}{{{{\cos }^2}\left( {\dfrac{{3x}}{2}} \right)}}\left( {\dfrac{3}{2}} \right) \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \left( {\dfrac{3}{2}} \right)\dfrac{1}{{\sin \left( {\dfrac{{3x}}{2}} \right)\cos \left( {\dfrac{{3x}}{2}} \right)}} \\
\]
Now, we know $2\sin \left( {\dfrac{a}{2}} \right)\cos \left( {\dfrac{a}{2}} \right) = \sin a$
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \left( {\dfrac{3}{2}} \right)\dfrac{1}{{\sin \left( {\dfrac{{3x}}{2}} \right)\cos \left( {\dfrac{{3x}}{2}} \right)}} = \dfrac{3}{{\sin 3x}} = 3\csc 3x,{\text{ }}\left( {\dfrac{1}{{\sin x}} = \csc x} \right)\]
So, this is the required differentiation.

Note: - In such a type of question the key concept is to remember the formula of differentiation of log, and also remember the half angle properties of sin and cosine, then simplify we will get the required answer.