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

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