Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 3\right)^{6} \sqrt{7 x + 6} e^{- x} \cos^{2}{\left(x \right)}}{\left(x + 5\right)^{8} \left(3 x + 3\right)^{3} \sin^{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 + 3\right)^{6} \sqrt{7 x + 6} e^{- x} \cos^{2}{\left(x \right)}}{\left(x + 5\right)^{8} \left(3 x + 3\right)^{3} \sin^{2}{\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 + 3 \right)} + \frac{\ln{\left(7 x + 6 \right)}}{2} + 2 \ln{\left(\cos{\left(x \right)} \right)}- x - 8 \ln{\left(x + 5 \right)} - 3 \ln{\left(3 x + 3 \right)} - 2 \ln{\left(\sin{\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{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{7}{2 \left(7 x + 6\right)} - \frac{9}{3 x + 3} - \frac{8}{x + 5} + \frac{6}{x + 3} 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{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{7}{2 \left(7 x + 6\right)} - \frac{9}{3 x + 3} - \frac{8}{x + 5} + \frac{6}{x + 3}\right)\left(\frac{\left(x + 3\right)^{6} \sqrt{7 x + 6} e^{- x} \cos^{2}{\left(x \right)}}{\left(x + 5\right)^{8} \left(3 x + 3\right)^{3} \sin^{2}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 2 \tan{\left(x \right)} + \frac{7}{2 \left(7 x + 6\right)} + \frac{6}{x + 3}-1 - \frac{2}{\tan{\left(x \right)}} - \frac{9}{3 x + 3} - \frac{8}{x + 5}\right)\left(\frac{\left(x + 3\right)^{6} \sqrt{7 x + 6} e^{- x} \cos^{2}{\left(x \right)}}{\left(x + 5\right)^{8} \left(3 x + 3\right)^{3} \sin^{2}{\left(x \right)}} \right)