Find the derivative of LaTeX: \displaystyle y = \frac{\sqrt{\left(2 x + 1\right)^{7}} \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)}}{\left(2 - 3 x\right)^{6} \left(x + 3\right)^{2}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\sqrt{\left(2 x + 1\right)^{7}} \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)}}{\left(2 - 3 x\right)^{6} \left(x + 3\right)^{2}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{7 \ln{\left(2 x + 1 \right)}}{2} + 2 \ln{\left(\sin{\left(x \right)} \right)} + 3 \ln{\left(\cos{\left(x \right)} \right)}- 6 \ln{\left(2 - 3 x \right)} - 2 \ln{\left(x + 3 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{3 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{7}{2 x + 1} - \frac{2}{x + 3} + \frac{18}{2 - 3 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{3 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{7}{2 x + 1} - \frac{2}{x + 3} + \frac{18}{2 - 3 x}\right)\left(\frac{\sqrt{\left(2 x + 1\right)^{7}} \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)}}{\left(2 - 3 x\right)^{6} \left(x + 3\right)^{2}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 3 \tan{\left(x \right)} + \frac{2}{\tan{\left(x \right)}} + \frac{7}{2 x + 1}- \frac{2}{x + 3} + \frac{18}{2 - 3 x}\right)\left(\frac{\sqrt{\left(2 x + 1\right)^{7}} \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)}}{\left(2 - 3 x\right)^{6} \left(x + 3\right)^{2}} \right)