Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 6\right)^{2} \left(8 x - 7\right)^{4} e^{x} \cos^{7}{\left(x \right)}}{\left(x - 2\right)^{6} \sin^{7}{\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)^{2} \left(8 x - 7\right)^{4} e^{x} \cos^{7}{\left(x \right)}}{\left(x - 2\right)^{6} \sin^{7}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 2 \ln{\left(x + 6 \right)} + 4 \ln{\left(8 x - 7 \right)} + 7 \ln{\left(\cos{\left(x \right)} \right)}- 6 \ln{\left(x - 2 \right)} - 7 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 - \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{32}{8 x - 7} + \frac{2}{x + 6} - \frac{6}{x - 2} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 - \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{32}{8 x - 7} + \frac{2}{x + 6} - \frac{6}{x - 2}\right)\left(\frac{\left(x + 6\right)^{2} \left(8 x - 7\right)^{4} e^{x} \cos^{7}{\left(x \right)}}{\left(x - 2\right)^{6} \sin^{7}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 7 \tan{\left(x \right)} + 1 + \frac{32}{8 x - 7} + \frac{2}{x + 6}- \frac{7}{\tan{\left(x \right)}} - \frac{6}{x - 2}\right)\left(\frac{\left(x + 6\right)^{2} \left(8 x - 7\right)^{4} e^{x} \cos^{7}{\left(x \right)}}{\left(x - 2\right)^{6} \sin^{7}{\left(x \right)}} \right)