Find the derivative of LaTeX: \displaystyle y = \frac{\left(7 - 6 x\right)^{8} \left(5 x - 6\right)^{8} e^{x}}{\left(x + 5\right)^{7} \cos^{6}{\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(7 - 6 x\right)^{8} \left(5 x - 6\right)^{8} e^{x}}{\left(x + 5\right)^{7} \cos^{6}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 8 \ln{\left(7 - 6 x \right)} + 8 \ln{\left(5 x - 6 \right)}- 7 \ln{\left(x + 5 \right)} - 6 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{6 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{40}{5 x - 6} - \frac{7}{x + 5} - \frac{48}{7 - 6 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{6 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{40}{5 x - 6} - \frac{7}{x + 5} - \frac{48}{7 - 6 x}\right)\left(\frac{\left(7 - 6 x\right)^{8} \left(5 x - 6\right)^{8} e^{x}}{\left(x + 5\right)^{7} \cos^{6}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{40}{5 x - 6} - \frac{48}{7 - 6 x}6 \tan{\left(x \right)} - \frac{7}{x + 5}\right)\left(\frac{\left(7 - 6 x\right)^{8} \left(5 x - 6\right)^{8} e^{x}}{\left(x + 5\right)^{7} \cos^{6}{\left(x \right)}} \right)