Question

Show that \(\frac{d}{{dx}}\ln \sqrt {\frac{{1 - \cos x}}{{1 + \cos x}}} = \csc x\).

Short answer

It is proved that \(\frac{d}{{dx}}\left( {\ln \sqrt {\frac{{1 - \cos x}}{{1 + \cos x}}} } \right) = \csc x\).

Step-by-step solution

Derivative of logarithmic functions

The derivative of a logarithmicfunctionis shown below:

1. \(\frac{d}{{dx}}\left( {{{\log }_b}x} \right) = \frac{1}{{x\ln b}}\)
2. \(\frac{d}{{dx}}\left( {\ln x} \right) = \frac{1}{x}\)
3. \(\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}}\) or \(\frac{d}{{dx}}\left( {\ln g\left( x \right)} \right) = \frac{{g'\left( x \right)}}{{g\left( x \right)}}\)

Show that \(\frac{d}{{dx}}\ln \sqrt {\frac{{1 - \cos x}}{{1 + \cos x}}}  = \csc x\)

Evaluate the derivative of the function as shown below:

\(\begin{aligned}{c}\frac{d}{{dx}}\left( {\ln \sqrt {\frac{{1 - \cos x}}{{1 + \cos x}}} } \right)&= \frac{d}{{dx}}\left( {\ln \sqrt {1 - \cos x} - \ln \sqrt {1 + \cos x} } \right)\\&= \frac{d}{{dx}}\left( {\frac{1}{2}\ln \left( {1 - \cos x} \right) - \frac{1}{2}\ln \left( {1 + \cos x} \right)} \right)\\&= \frac{1}{2} \cdot \frac{1}{{\left( {1 - \cos x} \right)}} \cdot \frac{d}{{dx}}\left( {1 - \cos x} \right) - \frac{1}{2} \cdot \frac{1}{{\left( {1 + \cos x} \right)}} \cdot \frac{d}{{dx}}\left( {1 + \cos x} \right)\\&= \frac{1}{2} \cdot \frac{1}{{\left( {1 - \cos x} \right)}} \cdot \left( {\sin x} \right) - \frac{1}{2} \cdot \frac{1}{{\left( {1 + \cos x} \right)}} \cdot \left( { - \sin x} \right)\end{aligned}\)

Simplify further,

\(\)\(\begin{aligned}{c}\frac{d}{{dx}}\left( {\ln \sqrt {\frac{{1 - \cos x}}{{1 + \cos x}}} } \right)&= \frac{1}{2}\left( {\frac{{\sin x}}{{1 - \cos x}} + \frac{{\sin x}}{{1 + \cos x}}} \right)\\&= \frac{1}{2}\left( {\frac{{\sin x\left( {1 + \cos x} \right) + \sin x\left( {1 - \cos x} \right)}}{{\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)}}} \right)\\&= \frac{1}{2}\left( {\frac{{\sin x + \sin x\cos x + \sin x - \sin x\cos x}}{{\left( {1 - {{\cos }^2}x} \right)}}} \right)\\&= \frac{1}{2}\left( {\frac{{2\sin x}}{{{{\sin }^2}x}}} \right)\\&= \frac{1}{{\sin x}}\\&= \csc x\end{aligned}\)

Thus, it is proved that \(\frac{d}{{dx}}\left( {\ln \sqrt {\frac{{1 - \cos x}}{{1 + \cos x}}} } \right) = \csc x\).