Find the derivative of LaTeX: \displaystyle y = \frac{\left(2 x + 7\right)^{2} \left(6 x - 3\right)^{7} e^{x} \sin^{8}{\left(x \right)}}{\sqrt{8 x + 1} \cos^{2}{\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(2 x + 7\right)^{2} \left(6 x - 3\right)^{7} e^{x} \sin^{8}{\left(x \right)}}{\sqrt{8 x + 1} \cos^{2}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 2 \ln{\left(2 x + 7 \right)} + 7 \ln{\left(6 x - 3 \right)} + 8 \ln{\left(\sin{\left(x \right)} \right)}- \frac{\ln{\left(8 x + 1 \right)}}{2} - 2 \ln{\left(\cos{\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{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{4}{8 x + 1} + \frac{42}{6 x - 3} + \frac{4}{2 x + 7} 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{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{4}{8 x + 1} + \frac{42}{6 x - 3} + \frac{4}{2 x + 7}\right)\left(\frac{\left(2 x + 7\right)^{2} \left(6 x - 3\right)^{7} e^{x} \sin^{8}{\left(x \right)}}{\sqrt{8 x + 1} \cos^{2}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{8}{\tan{\left(x \right)}} + \frac{42}{6 x - 3} + \frac{4}{2 x + 7}2 \tan{\left(x \right)} - \frac{4}{8 x + 1}\right)\left(\frac{\left(2 x + 7\right)^{2} \left(6 x - 3\right)^{7} e^{x} \sin^{8}{\left(x \right)}}{\sqrt{8 x + 1} \cos^{2}{\left(x \right)}} \right)