Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 6\right)^{3} \left(4 x + 8\right)^{3} e^{- x} \sin^{4}{\left(x \right)}}{\left(4 x + 7\right)^{5} \cos^{4}{\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)^{3} \left(4 x + 8\right)^{3} e^{- x} \sin^{4}{\left(x \right)}}{\left(4 x + 7\right)^{5} \cos^{4}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 3 \ln{\left(x + 6 \right)} + 3 \ln{\left(4 x + 8 \right)} + 4 \ln{\left(\sin{\left(x \right)} \right)}- x - 5 \ln{\left(4 x + 7 \right)} - 4 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{12}{4 x + 8} - \frac{20}{4 x + 7} + \frac{3}{x + 6} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{12}{4 x + 8} - \frac{20}{4 x + 7} + \frac{3}{x + 6}\right)\left(\frac{\left(x + 6\right)^{3} \left(4 x + 8\right)^{3} e^{- x} \sin^{4}{\left(x \right)}}{\left(4 x + 7\right)^{5} \cos^{4}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{4}{\tan{\left(x \right)}} + \frac{12}{4 x + 8} + \frac{3}{x + 6}4 \tan{\left(x \right)} - 1 - \frac{20}{4 x + 7}\right)\left(\frac{\left(x + 6\right)^{3} \left(4 x + 8\right)^{3} e^{- x} \sin^{4}{\left(x \right)}}{\left(4 x + 7\right)^{5} \cos^{4}{\left(x \right)}} \right)