Find \(y^\prime \) if \(y=\ln \left[ \dfrac{\left( 2x-1\right) ^{3}\sqrt{2x^{4}+1}}{x}\right]\!\).

In the remaining examples, we do not explicitly state the domain of a function containing a logarithm. Instead, we assume that the variable is restricted so all arguments for logarithmic functions are positive.

Solution Rather than attempting to use the Chain Rule, Quotient Rule, and Product Rule, we first simplify the right side by using properties of logarithms. \[ \begin{eqnarray*} y &=& \ln \left[ \dfrac{\left( 2x-1\right) ^{3}\sqrt{2x^{4}+1}}{x}\right] =\ln \left( 2x-1\right) ^{3}+\ln \sqrt{2x^{4}+1}-\ln x\\ &=& 3\ln \left( 2x-1\right) + \dfrac{1}{2}\ln ( 2x^{4}+1) -\ln x \end{eqnarray*} \]

Now we differentiate \(y.\) \[ \begin{eqnarray*} y^\prime &=&\dfrac{d}{dx}\left[ 3\ln \left( 2x-1\right) +\dfrac{1}{2}\ln ( 2x^{4}+1) -\ln x\right]\\ &=&\dfrac{d}{dx}\left[3\ln \left( 2x-1\right)\right] + \dfrac{d}{dx}\left[\dfrac{1}{2}\ln ( 2x^{4}+1)\right] -\dfrac{d}{dx}\ln x \\ &=&3\cdot \dfrac{2}{2x-1}+\dfrac{1}{2}\cdot \dfrac{8x^{3}}{2x^{4}+1}-\dfrac{1}{x}=\dfrac{6}{2x-1}+\dfrac{4x^{3}}{2x^{4}+1}-\dfrac{1}{x} \end{eqnarray*} \]