Find the derivative of LaTeX: \displaystyle y = \frac{\left(4 - x\right)^{3} \sqrt{\left(4 x + 5\right)^{5}} e^{x} \cos^{8}{\left(x \right)}}{\left(x - 3\right)^{7} \left(3 x + 3\right)^{6}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(4 - x\right)^{3} \sqrt{\left(4 x + 5\right)^{5}} e^{x} \cos^{8}{\left(x \right)}}{\left(x - 3\right)^{7} \left(3 x + 3\right)^{6}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 3 \ln{\left(4 - x \right)} + \frac{5 \ln{\left(4 x + 5 \right)}}{2} + 8 \ln{\left(\cos{\left(x \right)} \right)}- 7 \ln{\left(x - 3 \right)} - 6 \ln{\left(3 x + 3 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{8 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{10}{4 x + 5} - \frac{18}{3 x + 3} - \frac{7}{x - 3} - \frac{3}{4 - x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{8 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{10}{4 x + 5} - \frac{18}{3 x + 3} - \frac{7}{x - 3} - \frac{3}{4 - x}\right)\left(\frac{\left(4 - x\right)^{3} \sqrt{\left(4 x + 5\right)^{5}} e^{x} \cos^{8}{\left(x \right)}}{\left(x - 3\right)^{7} \left(3 x + 3\right)^{6}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 8 \tan{\left(x \right)} + 1 + \frac{10}{4 x + 5} - \frac{3}{4 - x}- \frac{18}{3 x + 3} - \frac{7}{x - 3}\right)\left(\frac{\left(4 - x\right)^{3} \sqrt{\left(4 x + 5\right)^{5}} e^{x} \cos^{8}{\left(x \right)}}{\left(x - 3\right)^{7} \left(3 x + 3\right)^{6}} \right)