Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 6\right)^{4} \left(8 x - 7\right)^{8} e^{x}}{\sqrt{2 x + 8} \sin^{7}{\left(x \right)} \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)^{4} \left(8 x - 7\right)^{8} e^{x}}{\sqrt{2 x + 8} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 4 \ln{\left(x + 6 \right)} + 8 \ln{\left(8 x - 7 \right)}- \frac{\ln{\left(2 x + 8 \right)}}{2} - 7 \ln{\left(\sin{\left(x \right)} \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{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{64}{8 x - 7} - \frac{1}{2 x + 8} + \frac{4}{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{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{64}{8 x - 7} - \frac{1}{2 x + 8} + \frac{4}{x + 6}\right)\left(\frac{\left(x + 6\right)^{4} \left(8 x - 7\right)^{8} e^{x}}{\sqrt{2 x + 8} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{64}{8 x - 7} + \frac{4}{x + 6}4 \tan{\left(x \right)} - \frac{7}{\tan{\left(x \right)}} - \frac{1}{2 x + 8}\right)\left(\frac{\left(x + 6\right)^{4} \left(8 x - 7\right)^{8} e^{x}}{\sqrt{2 x + 8} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}} \right)