Find the derivative of LaTeX: \displaystyle y = \frac{x^{2} \sqrt{\left(5 x + 5\right)^{7}} \sin^{3}{\left(x \right)}}{\left(- 9 x - 1\right)^{8} \left(5 x - 6\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{x^{2} \sqrt{\left(5 x + 5\right)^{7}} \sin^{3}{\left(x \right)}}{\left(- 9 x - 1\right)^{8} \left(5 x - 6\right)^{4}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 2 \ln{\left(x \right)} + \frac{7 \ln{\left(5 x + 5 \right)}}{2} + 3 \ln{\left(\sin{\left(x \right)} \right)}- 8 \ln{\left(- 9 x - 1 \right)} - 4 \ln{\left(5 x - 6 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{35}{2 \left(5 x + 5\right)} - \frac{20}{5 x - 6} + \frac{72}{- 9 x - 1} + \frac{2}{x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{35}{2 \left(5 x + 5\right)} - \frac{20}{5 x - 6} + \frac{72}{- 9 x - 1} + \frac{2}{x}\right)\left(\frac{x^{2} \sqrt{\left(5 x + 5\right)^{7}} \sin^{3}{\left(x \right)}}{\left(- 9 x - 1\right)^{8} \left(5 x - 6\right)^{4}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{3}{\tan{\left(x \right)}} + \frac{35}{2 \left(5 x + 5\right)} + \frac{2}{x}- \frac{20}{5 x - 6} + \frac{72}{- 9 x - 1}\right)\left(\frac{x^{2} \sqrt{\left(5 x + 5\right)^{7}} \sin^{3}{\left(x \right)}}{\left(- 9 x - 1\right)^{8} \left(5 x - 6\right)^{4}} \right)