Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 8\right)^{7} \sin^{8}{\left(x \right)}}{\left(4 x + 7\right)^{4} \sqrt{6 x + 2}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(x + 8\right)^{7} \sin^{8}{\left(x \right)}}{\left(4 x + 7\right)^{4} \sqrt{6 x + 2}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 7 \ln{\left(x + 8 \right)} + 8 \ln{\left(\sin{\left(x \right)} \right)}- 4 \ln{\left(4 x + 7 \right)} - \frac{\ln{\left(6 x + 2 \right)}}{2} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{3}{6 x + 2} - \frac{16}{4 x + 7} + \frac{7}{x + 8} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{3}{6 x + 2} - \frac{16}{4 x + 7} + \frac{7}{x + 8}\right)\left(\frac{\left(x + 8\right)^{7} \sin^{8}{\left(x \right)}}{\left(4 x + 7\right)^{4} \sqrt{6 x + 2}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{8}{\tan{\left(x \right)}} + \frac{7}{x + 8}- \frac{3}{6 x + 2} - \frac{16}{4 x + 7}\right)\left(\frac{\left(x + 8\right)^{7} \sin^{8}{\left(x \right)}}{\left(4 x + 7\right)^{4} \sqrt{6 x + 2}} \right)