Find the derivative of LaTeX: \displaystyle y = \frac{\left(5 - 6 x\right)^{5} \sin^{8}{\left(x \right)} \cos^{3}{\left(x \right)}}{\left(x + 5\right)^{7} \left(7 x - 5\right)^{4}}

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 - 6 x\right)^{5} \sin^{8}{\left(x \right)} \cos^{3}{\left(x \right)}}{\left(x + 5\right)^{7} \left(7 x - 5\right)^{4}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 5 \ln{\left(5 - 6 x \right)} + 8 \ln{\left(\sin{\left(x \right)} \right)} + 3 \ln{\left(\cos{\left(x \right)} \right)}- 7 \ln{\left(x + 5 \right)} - 4 \ln{\left(7 x - 5 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{3 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{28}{7 x - 5} - \frac{7}{x + 5} - \frac{30}{5 - 6 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{3 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{28}{7 x - 5} - \frac{7}{x + 5} - \frac{30}{5 - 6 x}\right)\left(\frac{\left(5 - 6 x\right)^{5} \sin^{8}{\left(x \right)} \cos^{3}{\left(x \right)}}{\left(x + 5\right)^{7} \left(7 x - 5\right)^{4}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 3 \tan{\left(x \right)} + \frac{8}{\tan{\left(x \right)}} - \frac{30}{5 - 6 x}- \frac{28}{7 x - 5} - \frac{7}{x + 5}\right)\left(\frac{\left(5 - 6 x\right)^{5} \sin^{8}{\left(x \right)} \cos^{3}{\left(x \right)}}{\left(x + 5\right)^{7} \left(7 x - 5\right)^{4}} \right)