Find $y^\prime$ if $y=\dfrac{x^{2}\sqrt{5x+1}}{( 3x-2) ^{3}}$.

Solution It is easier to find $y^\prime$ if we take the natural logarithm of each side before differentiating. That is, we write $$\ln y=\ln \left[ \dfrac{x^{2}\sqrt{5x+1}}{( 3x-2) ^{3}}\right]$$

and simplify the equation using properties of logarithms. $$\begin{eqnarray*} \ln y &=&\ln [ x^{2}\sqrt{5x+1}] -\ln ( 3x-2) ^{3}=\ln x^{2}+\ln (5x+1) ^{1/2}-\ln ( 3x-2) ^{3} \\ &=&2\ln x+\dfrac{1}{2}\ln (5x+1) -3\ln ( 3x-2) \end{eqnarray*}$$

To find $y^\prime$, we use implicit differentiation. $$\begin{eqnarray*} \dfrac{d}{dx}\ln y&=&\dfrac{d}{dx}\left[ 2\ln x+\dfrac{1}{2}\ln (5x+1) -3\ln (3x-2) \right]\\ \dfrac{y^\prime}{y}&=&\dfrac{d}{dx}(2\ln x) + \dfrac{d}{dx} \left[\dfrac{1}{2}\ln (5x+1)\right]-\dfrac{d}{dx}[3\ln(3x-2)]\\ &=&\dfrac{2}{x}+\dfrac{5}{2 ( 5x+1) }-\dfrac{9}{3x-2} \\ y^\prime &=&y\!\left[ \dfrac{2}{x}+\dfrac{5}{2 ( 5x+1) }-\dfrac{9}{3x-2}\right] \!=\!\left[ \dfrac{x^{2}\sqrt{5x+1}}{( 3x-2) ^{3}}\right]\!\! \left[ \dfrac{2}{x}+\dfrac{5}{2(5x+1) }-\dfrac{9}{3x-2}\right] \end{eqnarray*}$$