Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 6\right)^{4} \left(2 x - 7\right)^{7} e^{- x} \sin^{8}{\left(x \right)}}{\left(3 x - 3\right)^{4} \sqrt{7 x + 5} \cos^{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{\left(x - 6\right)^{4} \left(2 x - 7\right)^{7} e^{- x} \sin^{8}{\left(x \right)}}{\left(3 x - 3\right)^{4} \sqrt{7 x + 5} \cos^{5}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(x - 6 \right)} + 7 \ln{\left(2 x - 7 \right)} + 8 \ln{\left(\sin{\left(x \right)} \right)}- x - 4 \ln{\left(3 x - 3 \right)} - \frac{\ln{\left(7 x + 5 \right)}}{2} - 5 \ln{\left(\cos{\left(x \right)} \right)} 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{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{7}{2 \left(7 x + 5\right)} - \frac{12}{3 x - 3} + \frac{14}{2 x - 7} + \frac{4}{x - 6} 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{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{7}{2 \left(7 x + 5\right)} - \frac{12}{3 x - 3} + \frac{14}{2 x - 7} + \frac{4}{x - 6}\right)\left(\frac{\left(x - 6\right)^{4} \left(2 x - 7\right)^{7} e^{- x} \sin^{8}{\left(x \right)}}{\left(3 x - 3\right)^{4} \sqrt{7 x + 5} \cos^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{8}{\tan{\left(x \right)}} + \frac{14}{2 x - 7} + \frac{4}{x - 6}5 \tan{\left(x \right)} - 1 - \frac{7}{2 \left(7 x + 5\right)} - \frac{12}{3 x - 3}\right)\left(\frac{\left(x - 6\right)^{4} \left(2 x - 7\right)^{7} e^{- x} \sin^{8}{\left(x \right)}}{\left(3 x - 3\right)^{4} \sqrt{7 x + 5} \cos^{5}{\left(x \right)}} \right)