Find the derivative of LaTeX: \displaystyle y = \frac{\left(4 x - 1\right)^{3} \sqrt{\left(x + 5\right)^{5}} \cos^{2}{\left(x \right)}}{\left(x - 3\right)^{8} \sin^{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(4 x - 1\right)^{3} \sqrt{\left(x + 5\right)^{5}} \cos^{2}{\left(x \right)}}{\left(x - 3\right)^{8} \sin^{5}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{5 \ln{\left(x + 5 \right)}}{2} + 3 \ln{\left(4 x - 1 \right)} + 2 \ln{\left(\cos{\left(x \right)} \right)}- 8 \ln{\left(x - 3 \right)} - 5 \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)}} - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{12}{4 x - 1} + \frac{5}{2 \left(x + 5\right)} - \frac{8}{x - 3} 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)}} - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{12}{4 x - 1} + \frac{5}{2 \left(x + 5\right)} - \frac{8}{x - 3}\right)\left(\frac{\left(4 x - 1\right)^{3} \sqrt{\left(x + 5\right)^{5}} \cos^{2}{\left(x \right)}}{\left(x - 3\right)^{8} \sin^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 2 \tan{\left(x \right)} + \frac{12}{4 x - 1} + \frac{5}{2 \left(x + 5\right)}- \frac{5}{\tan{\left(x \right)}} - \frac{8}{x - 3}\right)\left(\frac{\left(4 x - 1\right)^{3} \sqrt{\left(x + 5\right)^{5}} \cos^{2}{\left(x \right)}}{\left(x - 3\right)^{8} \sin^{5}{\left(x \right)}} \right)