Find the derivative of LaTeX: \displaystyle y = \frac{\left(9 - x\right)^{8} \sin^{4}{\left(x \right)}}{\left(x - 5\right)^{3} \left(9 x + 8\right)^{5} \cos^{5}{\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(9 - x\right)^{8} \sin^{4}{\left(x \right)}}{\left(x - 5\right)^{3} \left(9 x + 8\right)^{5} \cos^{5}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 8 \ln{\left(9 - x \right)} + 4 \ln{\left(\sin{\left(x \right)} \right)}- 3 \ln{\left(x - 5 \right)} - 5 \ln{\left(9 x + 8 \right)} - 5 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{45}{9 x + 8} - \frac{3}{x - 5} - \frac{8}{9 - x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{45}{9 x + 8} - \frac{3}{x - 5} - \frac{8}{9 - x}\right)\left(\frac{\left(9 - x\right)^{8} \sin^{4}{\left(x \right)}}{\left(x - 5\right)^{3} \left(9 x + 8\right)^{5} \cos^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{4}{\tan{\left(x \right)}} - \frac{8}{9 - x}5 \tan{\left(x \right)} - \frac{45}{9 x + 8} - \frac{3}{x - 5}\right)\left(\frac{\left(9 - x\right)^{8} \sin^{4}{\left(x \right)}}{\left(x - 5\right)^{3} \left(9 x + 8\right)^{5} \cos^{5}{\left(x \right)}} \right)