Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 5 x - 3\right)^{5} e^{x} \sin^{8}{\left(x \right)}}{\left(2 x + 3\right)^{4} \left(5 x + 8\right)^{5} \sqrt{\left(7 x + 8\right)^{7}}}

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 - 3\right)^{5} e^{x} \sin^{8}{\left(x \right)}}{\left(2 x + 3\right)^{4} \left(5 x + 8\right)^{5} \sqrt{\left(7 x + 8\right)^{7}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 5 \ln{\left(- 5 x - 3 \right)} + 8 \ln{\left(\sin{\left(x \right)} \right)}- 4 \ln{\left(2 x + 3 \right)} - 5 \ln{\left(5 x + 8 \right)} - \frac{7 \ln{\left(7 x + 8 \right)}}{2} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = 1 + \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{49}{2 \left(7 x + 8\right)} - \frac{25}{5 x + 8} - \frac{8}{2 x + 3} - \frac{25}{- 5 x - 3} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(1 + \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{49}{2 \left(7 x + 8\right)} - \frac{25}{5 x + 8} - \frac{8}{2 x + 3} - \frac{25}{- 5 x - 3}\right)\left(\frac{\left(- 5 x - 3\right)^{5} e^{x} \sin^{8}{\left(x \right)}}{\left(2 x + 3\right)^{4} \left(5 x + 8\right)^{5} \sqrt{\left(7 x + 8\right)^{7}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{8}{\tan{\left(x \right)}} - \frac{25}{- 5 x - 3}- \frac{49}{2 \left(7 x + 8\right)} - \frac{25}{5 x + 8} - \frac{8}{2 x + 3}\right)\left(\frac{\left(- 5 x - 3\right)^{5} e^{x} \sin^{8}{\left(x \right)}}{\left(2 x + 3\right)^{4} \left(5 x + 8\right)^{5} \sqrt{\left(7 x + 8\right)^{7}}} \right)