Find the derivative of LaTeX: \displaystyle y = \frac{\left(4 - 5 x\right)^{4} \left(x - 9\right)^{3} e^{x} \cos^{2}{\left(x \right)}}{\left(x - 2\right)^{3} \sqrt{x + 5} \sin^{3}{\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(4 - 5 x\right)^{4} \left(x - 9\right)^{3} e^{x} \cos^{2}{\left(x \right)}}{\left(x - 2\right)^{3} \sqrt{x + 5} \sin^{3}{\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(4 - 5 x \right)} + 3 \ln{\left(x - 9 \right)} + 2 \ln{\left(\cos{\left(x \right)} \right)}- 3 \ln{\left(x - 2 \right)} - \frac{\ln{\left(x + 5 \right)}}{2} - 3 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 - \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{1}{2 \left(x + 5\right)} - \frac{3}{x - 2} + \frac{3}{x - 9} - \frac{20}{4 - 5 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 - \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{1}{2 \left(x + 5\right)} - \frac{3}{x - 2} + \frac{3}{x - 9} - \frac{20}{4 - 5 x}\right)\left(\frac{\left(4 - 5 x\right)^{4} \left(x - 9\right)^{3} e^{x} \cos^{2}{\left(x \right)}}{\left(x - 2\right)^{3} \sqrt{x + 5} \sin^{3}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 2 \tan{\left(x \right)} + 1 + \frac{3}{x - 9} - \frac{20}{4 - 5 x}- \frac{3}{\tan{\left(x \right)}} - \frac{1}{2 \left(x + 5\right)} - \frac{3}{x - 2}\right)\left(\frac{\left(4 - 5 x\right)^{4} \left(x - 9\right)^{3} e^{x} \cos^{2}{\left(x \right)}}{\left(x - 2\right)^{3} \sqrt{x + 5} \sin^{3}{\left(x \right)}} \right)