Find the derivative of LaTeX: \displaystyle y = \frac{729 x^{6} \left(x + 3\right)^{7} e^{- x}}{\left(- 8 x - 7\right)^{8} \left(4 x + 7\right)^{4} \sin^{5}{\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{729 x^{6} \left(x + 3\right)^{7} e^{- x}}{\left(- 8 x - 7\right)^{8} \left(4 x + 7\right)^{4} \sin^{5}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 6 \ln{\left(x \right)} + 7 \ln{\left(x + 3 \right)} + 6 \ln{\left(3 \right)}- x - 8 \ln{\left(- 8 x - 7 \right)} - 4 \ln{\left(4 x + 7 \right)} - 5 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{16}{4 x + 7} + \frac{7}{x + 3} + \frac{64}{- 8 x - 7} + \frac{6}{x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{16}{4 x + 7} + \frac{7}{x + 3} + \frac{64}{- 8 x - 7} + \frac{6}{x}\right)\left(\frac{729 x^{6} \left(x + 3\right)^{7} e^{- x}}{\left(- 8 x - 7\right)^{8} \left(4 x + 7\right)^{4} \sin^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{7}{x + 3} + \frac{6}{x}-1 - \frac{5}{\tan{\left(x \right)}} - \frac{16}{4 x + 7} + \frac{64}{- 8 x - 7}\right)\left(\frac{729 x^{6} \left(x + 3\right)^{7} e^{- x}}{\left(- 8 x - 7\right)^{8} \left(4 x + 7\right)^{4} \sin^{5}{\left(x \right)}} \right)