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

If y=log⁡1−cos⁡3x1+cos⁡3xy = \log \sqrt {\dfrac{{1 - \cos 3x}}{{1 + \cos 3x}}} , find dydx.\dfrac{{dy}}{{dx}}.

Explanation

Solution

Hint: Convert trigonometric terms in their half angles.

As we know the differentiation of log⁡(ax+b)=1ax+b(ddx(ax+b))\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
dydx=11−cos⁡3x1+cos⁡3x(ddx(1−cos⁡3x1+cos⁡3x))..............(1)\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(1−cos⁡ax)=2sin⁡2(ax2), (1+cos⁡ax)=2cos⁡2(ax2)\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
dydx=12sin⁡2(3x2)2cos⁡2(3x2)(ddx(2sin⁡2(3x2)2cos⁡2(3x2))) sin⁡xcos⁡x=tan⁡x ⇒dydx=1tan⁡2(3x2)(ddx(tan⁡2(3x2))) ⇒dydx=1tan⁡(3x2)ddx(tan⁡(3x2))  \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\tan xdifferentiation is sec⁡2x{\sec ^2}x

⇒dydx=1tan⁡(3x2)sec⁡2(3x2)(ddx3x2) ⇒dydx=1sin⁡(3x2)cos⁡(3x2)1cos⁡2(3x2)(32) ⇒dydx=(32)1sin⁡(3x2)cos⁡(3x2)  \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 2sin⁡(a2)cos⁡(a2)=sin⁡a2\sin \left( {\dfrac{a}{2}} \right)\cos \left( {\dfrac{a}{2}} \right) = \sin a
⇒dydx=(32)1sin⁡(3x2)cos⁡(3x2)=3sin⁡3x=3csc⁡3x, (1sin⁡x=csc⁡x)\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.