Find the derivative of LaTeX: \displaystyle y = \frac{\left(5 x + 1\right)^{5} \sin^{5}{\left(x \right)}}{\left(7 - x\right)^{8} \left(x + 1\right)^{6}}

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 x + 1\right)^{5} \sin^{5}{\left(x \right)}}{\left(7 - x\right)^{8} \left(x + 1\right)^{6}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 5 \ln{\left(5 x + 1 \right)} + 5 \ln{\left(\sin{\left(x \right)} \right)}- 8 \ln{\left(7 - x \right)} - 6 \ln{\left(x + 1 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{25}{5 x + 1} - \frac{6}{x + 1} + \frac{8}{7 - x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{25}{5 x + 1} - \frac{6}{x + 1} + \frac{8}{7 - x}\right)\left(\frac{\left(5 x + 1\right)^{5} \sin^{5}{\left(x \right)}}{\left(7 - x\right)^{8} \left(x + 1\right)^{6}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{5}{\tan{\left(x \right)}} + \frac{25}{5 x + 1}- \frac{6}{x + 1} + \frac{8}{7 - x}\right)\left(\frac{\left(5 x + 1\right)^{5} \sin^{5}{\left(x \right)}}{\left(7 - x\right)^{8} \left(x + 1\right)^{6}} \right)