Find the derivative of LaTeX: \displaystyle y = \frac{\sqrt{x + 1} \left(5 x + 8\right)^{4}}{\sin^{8}{\left(x \right)} \cos^{8}{\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{\sqrt{x + 1} \left(5 x + 8\right)^{4}}{\sin^{8}{\left(x \right)} \cos^{8}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{\ln{\left(x + 1 \right)}}{2} + 4 \ln{\left(5 x + 8 \right)}- 8 \ln{\left(\sin{\left(x \right)} \right)} - 8 \ln{\left(\cos{\left(x \right)} \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)}} - \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{20}{5 x + 8} + \frac{1}{2 \left(x + 1\right)} 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)}} - \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{20}{5 x + 8} + \frac{1}{2 \left(x + 1\right)}\right)\left(\frac{\sqrt{x + 1} \left(5 x + 8\right)^{4}}{\sin^{8}{\left(x \right)} \cos^{8}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{20}{5 x + 8} + \frac{1}{2 \left(x + 1\right)}8 \tan{\left(x \right)} - \frac{8}{\tan{\left(x \right)}}\right)\left(\frac{\sqrt{x + 1} \left(5 x + 8\right)^{4}}{\sin^{8}{\left(x \right)} \cos^{8}{\left(x \right)}} \right)