Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 4\right)^{7} \left(3 x - 3\right)^{3} e^{- x} \sin^{7}{\left(x \right)}}{\left(x + 1\right)^{6} \left(6 x - 4\right)^{2} \cos^{2}{\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 + 4\right)^{7} \left(3 x - 3\right)^{3} e^{- x} \sin^{7}{\left(x \right)}}{\left(x + 1\right)^{6} \left(6 x - 4\right)^{2} \cos^{2}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 7 \ln{\left(x + 4 \right)} + 3 \ln{\left(3 x - 3 \right)} + 7 \ln{\left(\sin{\left(x \right)} \right)}- x - 6 \ln{\left(x + 1 \right)} - 2 \ln{\left(6 x - 4 \right)} - 2 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{12}{6 x - 4} + \frac{9}{3 x - 3} + \frac{7}{x + 4} - \frac{6}{x + 1} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{12}{6 x - 4} + \frac{9}{3 x - 3} + \frac{7}{x + 4} - \frac{6}{x + 1}\right)\left(\frac{\left(x + 4\right)^{7} \left(3 x - 3\right)^{3} e^{- x} \sin^{7}{\left(x \right)}}{\left(x + 1\right)^{6} \left(6 x - 4\right)^{2} \cos^{2}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{7}{\tan{\left(x \right)}} + \frac{9}{3 x - 3} + \frac{7}{x + 4}2 \tan{\left(x \right)} - 1 - \frac{12}{6 x - 4} - \frac{6}{x + 1}\right)\left(\frac{\left(x + 4\right)^{7} \left(3 x - 3\right)^{3} e^{- x} \sin^{7}{\left(x \right)}}{\left(x + 1\right)^{6} \left(6 x - 4\right)^{2} \cos^{2}{\left(x \right)}} \right)