Find the derivative of LaTeX: \displaystyle y = \frac{\left(- x - 5\right)^{7} e^{x} \sin^{2}{\left(x \right)}}{\left(x + 3\right)^{4} \left(2 x + 1\right)^{5} \cos^{3}{\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(- x - 5\right)^{7} e^{x} \sin^{2}{\left(x \right)}}{\left(x + 3\right)^{4} \left(2 x + 1\right)^{5} \cos^{3}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 7 \ln{\left(- x - 5 \right)} + 2 \ln{\left(\sin{\left(x \right)} \right)}- 4 \ln{\left(x + 3 \right)} - 5 \ln{\left(2 x + 1 \right)} - 3 \ln{\left(\cos{\left(x \right)} \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)}} + 1 + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{10}{2 x + 1} - \frac{4}{x + 3} - \frac{7}{- x - 5} 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)}} + 1 + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{10}{2 x + 1} - \frac{4}{x + 3} - \frac{7}{- x - 5}\right)\left(\frac{\left(- x - 5\right)^{7} e^{x} \sin^{2}{\left(x \right)}}{\left(x + 3\right)^{4} \left(2 x + 1\right)^{5} \cos^{3}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{2}{\tan{\left(x \right)}} - \frac{7}{- x - 5}3 \tan{\left(x \right)} - \frac{10}{2 x + 1} - \frac{4}{x + 3}\right)\left(\frac{\left(- x - 5\right)^{7} e^{x} \sin^{2}{\left(x \right)}}{\left(x + 3\right)^{4} \left(2 x + 1\right)^{5} \cos^{3}{\left(x \right)}} \right)