Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 6 x - 4\right)^{2} \sqrt{\left(6 x + 5\right)^{5}} e^{x}}{\left(3 - 5 x\right)^{3} \sin^{7}{\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(- 6 x - 4\right)^{2} \sqrt{\left(6 x + 5\right)^{5}} e^{x}}{\left(3 - 5 x\right)^{3} \sin^{7}{\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(- 6 x - 4 \right)} + \frac{5 \ln{\left(6 x + 5 \right)}}{2}- 3 \ln{\left(3 - 5 x \right)} - 7 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = 1 - \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{15}{6 x + 5} - \frac{12}{- 6 x - 4} + \frac{15}{3 - 5 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(1 - \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{15}{6 x + 5} - \frac{12}{- 6 x - 4} + \frac{15}{3 - 5 x}\right)\left(\frac{\left(- 6 x - 4\right)^{2} \sqrt{\left(6 x + 5\right)^{5}} e^{x}}{\left(3 - 5 x\right)^{3} \sin^{7}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{15}{6 x + 5} - \frac{12}{- 6 x - 4}- \frac{7}{\tan{\left(x \right)}} + \frac{15}{3 - 5 x}\right)\left(\frac{\left(- 6 x - 4\right)^{2} \sqrt{\left(6 x + 5\right)^{5}} e^{x}}{\left(3 - 5 x\right)^{3} \sin^{7}{\left(x \right)}} \right)