Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 5 x - 1\right)^{5} \sqrt{\left(2 x + 5\right)^{7}} e^{x}}{\left(x + 4\right)^{5} \sin^{5}{\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(- 5 x - 1\right)^{5} \sqrt{\left(2 x + 5\right)^{7}} e^{x}}{\left(x + 4\right)^{5} \sin^{5}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 5 \ln{\left(- 5 x - 1 \right)} + \frac{7 \ln{\left(2 x + 5 \right)}}{2}- 5 \ln{\left(x + 4 \right)} - 5 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = 1 - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{7}{2 x + 5} - \frac{5}{x + 4} - \frac{25}{- 5 x - 1} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(1 - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{7}{2 x + 5} - \frac{5}{x + 4} - \frac{25}{- 5 x - 1}\right)\left(\frac{\left(- 5 x - 1\right)^{5} \sqrt{\left(2 x + 5\right)^{7}} e^{x}}{\left(x + 4\right)^{5} \sin^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{7}{2 x + 5} - \frac{25}{- 5 x - 1}- \frac{5}{\tan{\left(x \right)}} - \frac{5}{x + 4}\right)\left(\frac{\left(- 5 x - 1\right)^{5} \sqrt{\left(2 x + 5\right)^{7}} e^{x}}{\left(x + 4\right)^{5} \sin^{5}{\left(x \right)}} \right)