Question

Differentiate the function.

18. \(g\left( t \right) = \ln \frac{{t{{\left( {{t^2} + 1} \right)}^4}}}{{\sqrt(3){{2t - 1}}}}\)

Short answer

The derivative of the function is \(g'\left( t \right) = \frac{1}{t} + \frac{{8t}}{{{t^2} + 1}} - \frac{2}{{3\left( {2t - 1} \right)}}\)

Step-by-step solution

Use the derivative of logarithmic function

Rule 2: The derivative of \(\ln x\) is,

\(\frac{d}{{dx}}\left( {\ln x} \right) = \frac{1}{x}\)

Evaluating the derivative of given function

Simplify the given function as follows:

\(\begin{aligned}{c}g\left( t \right)&= \ln \frac{{t{{\left( {{t^2} + 1} \right)}^4}}}{{\sqrt(3){{2t - 1}}}}\\g\left( t \right)&= \ln \frac{{t{{\left( {{t^2} + 1} \right)}^4}}}{{{{\left( {2t - 1} \right)}^{1/3}}}}\\g\left( t \right)&= \ln t\left( {{t^2} + 1} \right) - \ln {\left( {2t - 1} \right)^{1/3}}\\g\left( t \right)&= \ln t + \ln {\left( {{t^2} + 1} \right)^4} - \frac{1}{3}\ln \left( {2t - 1} \right)\\\end{aligned}\)

Differentiating \(g\left( t \right)\)with respect to t,

\(\begin{aligned}{c}\frac{d}{{dt}}g\left( t \right)&= \frac{d}{{dt}}\left( {\ln t} \right) + 4\frac{{d\ln \left( {{t^2} + 1} \right)}}{{dt}} - \frac{1}{3}\frac{d}{{dt}}\ln \left( {2t - 1} \right)\\{g^{'}}\left( t \right)&= \frac{1}{t} + 4\frac{{d\ln \left( {{t^2} + 1} \right)}}{{dt}} \times \frac{{d\left( {{t^2} + 1} \right)}}{{dt}} - \frac{1}{3}\frac{{d\log \left( {2t - 1} \right)}}{{d\left( {2t - 1} \right)}} \times \frac{{d\left( {2t - 1} \right)}}{{dt}}\\{g^{'}}\left( t \right)&= \frac{1}{t} + \frac{4}{{\left( {{t^2} + 1} \right)}} \times \left( {2t + 0} \right) - \frac{1}{{3\left( {2t - 1} \right)}} \times \left( {2 - 0} \right)\\{g^{'}}\left( t \right)&= \frac{1}{t} + \frac{{8t}}{{{t^2} + 1}} - \frac{2}{{3\left( {2t - 1} \right)}}\end{aligned}\)

Thus, the value of the derivative is \(g'\left( t \right) = \frac{1}{t} + \frac{{8t}}{{{t^2} + 1}} - \frac{2}{{3\left( {2t - 1} \right)}}\).