Find the derivative of LaTeX: \displaystyle y = \frac{\left(4 - x\right)^{4} \left(x - 8\right)^{6} \sqrt{\left(x + 6\right)^{7}} e^{- x}}{\left(7 x - 7\right)^{5} \sin^{7}{\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{\left(4 - x\right)^{4} \left(x - 8\right)^{6} \sqrt{\left(x + 6\right)^{7}} e^{- x}}{\left(7 x - 7\right)^{5} \sin^{7}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(4 - x \right)} + 6 \ln{\left(x - 8 \right)} + \frac{7 \ln{\left(x + 6 \right)}}{2}- x - 5 \ln{\left(7 x - 7 \right)} - 7 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 - \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{35}{7 x - 7} + \frac{7}{2 \left(x + 6\right)} + \frac{6}{x - 8} - \frac{4}{4 - x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 - \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{35}{7 x - 7} + \frac{7}{2 \left(x + 6\right)} + \frac{6}{x - 8} - \frac{4}{4 - x}\right)\left(\frac{\left(4 - x\right)^{4} \left(x - 8\right)^{6} \sqrt{\left(x + 6\right)^{7}} e^{- x}}{\left(7 x - 7\right)^{5} \sin^{7}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{7}{2 \left(x + 6\right)} + \frac{6}{x - 8} - \frac{4}{4 - x}-1 - \frac{7}{\tan{\left(x \right)}} - \frac{35}{7 x - 7}\right)\left(\frac{\left(4 - x\right)^{4} \left(x - 8\right)^{6} \sqrt{\left(x + 6\right)^{7}} e^{- x}}{\left(7 x - 7\right)^{5} \sin^{7}{\left(x \right)}} \right)