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

Answer

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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.

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.

Last updated date: 26th Sep 2023

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