Find the derivative of LaTeX: \displaystyle y = \frac{e^{- x} \sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)}}{\left(4 - x\right)^{4} \left(3 x + 4\right)^{8} \sqrt{\left(4 x + 2\right)^{5}}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{e^{- x} \sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)}}{\left(4 - x\right)^{4} \left(3 x + 4\right)^{8} \sqrt{\left(4 x + 2\right)^{5}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 2 \ln{\left(\sin{\left(x \right)} \right)} + 5 \ln{\left(\cos{\left(x \right)} \right)}- x - 4 \ln{\left(4 - x \right)} - 8 \ln{\left(3 x + 4 \right)} - \frac{5 \ln{\left(4 x + 2 \right)}}{2} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{10}{4 x + 2} - \frac{24}{3 x + 4} + \frac{4}{4 - x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{10}{4 x + 2} - \frac{24}{3 x + 4} + \frac{4}{4 - x}\right)\left(\frac{e^{- x} \sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)}}{\left(4 - x\right)^{4} \left(3 x + 4\right)^{8} \sqrt{\left(4 x + 2\right)^{5}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 5 \tan{\left(x \right)} + \frac{2}{\tan{\left(x \right)}}-1 - \frac{10}{4 x + 2} - \frac{24}{3 x + 4} + \frac{4}{4 - x}\right)\left(\frac{e^{- x} \sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)}}{\left(4 - x\right)^{4} \left(3 x + 4\right)^{8} \sqrt{\left(4 x + 2\right)^{5}}} \right)