Find the derivative of LaTeX: \displaystyle y = \frac{81 x^{4} \left(2 - 3 x\right)^{8} \left(8 - 4 x\right)^{8} e^{x}}{\left(x + 4\right)^{4} \sin^{3}{\left(x \right)}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{81 x^{4} \left(2 - 3 x\right)^{8} \left(8 - 4 x\right)^{8} e^{x}}{\left(x + 4\right)^{4} \sin^{3}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 4 \ln{\left(x \right)} + 8 \ln{\left(2 - 3 x \right)} + 8 \ln{\left(8 - 4 x \right)} + 4 \ln{\left(3 \right)}- 4 \ln{\left(x + 4 \right)} - 3 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = 1 - \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{4}{x + 4} - \frac{32}{8 - 4 x} - \frac{24}{2 - 3 x} + \frac{4}{x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(1 - \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{4}{x + 4} - \frac{32}{8 - 4 x} - \frac{24}{2 - 3 x} + \frac{4}{x}\right)\left(\frac{81 x^{4} \left(2 - 3 x\right)^{8} \left(8 - 4 x\right)^{8} e^{x}}{\left(x + 4\right)^{4} \sin^{3}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 - \frac{32}{8 - 4 x} - \frac{24}{2 - 3 x} + \frac{4}{x}- \frac{3}{\tan{\left(x \right)}} - \frac{4}{x + 4}\right)\left(\frac{81 x^{4} \left(2 - 3 x\right)^{8} \left(8 - 4 x\right)^{8} e^{x}}{\left(x + 4\right)^{4} \sin^{3}{\left(x \right)}} \right)