Question

Differentiate the function.

24.\(y = \ln \sqrt {\frac{{1 + 2x}}{{1 - 2x}}} \)

Short answer

The derivative of the function is \(y' = \frac{1}{{1 + 2x}} + \frac{1}{{1 - 2x}}\).

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 \(\frac{d}{{dx}}\left( {\ln g\left( x \right)} \right) = \frac{{g'\left( x \right)}}{{g\left( x \right)}}\)

Differentiate the function

Rewrite the function as shown below:

\(\begin{aligned}{c}y&= \ln \sqrt {\frac{{1 + 2x}}{{1 - 2x}}} \\&= \ln \sqrt {1 + 2x} - \ln \sqrt {1 - 2x} \\&= \frac{1}{2}\ln \left( {1 + 2x} \right) - \frac{1}{2}\ln \left( {1 - 2x} \right)\end{aligned}\)

Use the formula to differentiate the function as shown below:

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

Thus, the derivative of the function is \(y' = \frac{1}{{1 + 2x}} + \frac{1}{{1 - 2x}}\).