Find the derivative of LaTeX: \displaystyle y = \frac{\left(5 - 8 x\right)^{2} \sqrt{\left(4 x + 8\right)^{3}} \cos^{5}{\left(x \right)}}{\left(x + 5\right)^{2} \left(5 x - 3\right)^{3}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(5 - 8 x\right)^{2} \sqrt{\left(4 x + 8\right)^{3}} \cos^{5}{\left(x \right)}}{\left(x + 5\right)^{2} \left(5 x - 3\right)^{3}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 2 \ln{\left(5 - 8 x \right)} + \frac{3 \ln{\left(4 x + 8 \right)}}{2} + 5 \ln{\left(\cos{\left(x \right)} \right)}- 2 \ln{\left(x + 5 \right)} - 3 \ln{\left(5 x - 3 \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{15}{5 x - 3} + \frac{6}{4 x + 8} - \frac{2}{x + 5} - \frac{16}{5 - 8 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{15}{5 x - 3} + \frac{6}{4 x + 8} - \frac{2}{x + 5} - \frac{16}{5 - 8 x}\right)\left(\frac{\left(5 - 8 x\right)^{2} \sqrt{\left(4 x + 8\right)^{3}} \cos^{5}{\left(x \right)}}{\left(x + 5\right)^{2} \left(5 x - 3\right)^{3}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 5 \tan{\left(x \right)} + \frac{6}{4 x + 8} - \frac{16}{5 - 8 x}- \frac{15}{5 x - 3} - \frac{2}{x + 5}\right)\left(\frac{\left(5 - 8 x\right)^{2} \sqrt{\left(4 x + 8\right)^{3}} \cos^{5}{\left(x \right)}}{\left(x + 5\right)^{2} \left(5 x - 3\right)^{3}} \right)