Find the derivative of $LaTeX: \displaystyle y = - \frac{\left(x - 4\right)^{8} \sin^{4}{\left(x \right)}}{78125 x^{7} \left(x + 3\right)^{3} \sqrt{\left(4 x + 6\right)^{3}}}$

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: $LaTeX: \ln(y) = \ln{\left(- \frac{\left(x - 4\right)^{8} \sin^{4}{\left(x \right)}}{78125 x^{7} \left(x + 3\right)^{3} \sqrt{\left(4 x + 6\right)^{3}}} \right)}$ Expanding the right hand side using the product and quotient properties of logarithms gives: $LaTeX: \ln(y) = 8 \ln{\left(x - 4 \right)} + 4 \ln{\left(\sin{\left(x \right)} \right)}- 7 \ln{\left(x \right)} - 3 \ln{\left(x + 3 \right)} - \frac{3 \ln{\left(4 x + 6 \right)}}{2} - 7 \ln{\left(5 \right)} - i \pi$ Taking the derivative on both sides of the equation yields: $LaTeX: \frac{y'}{y} = \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{6}{4 x + 6} - \frac{3}{x + 3} + \frac{8}{x - 4} - \frac{7}{x}$ Solving for $LaTeX: \displaystyle y'$ and substituting out y using the original equation gives $LaTeX: y' = \left(\frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{6}{4 x + 6} - \frac{3}{x + 3} + \frac{8}{x - 4} - \frac{7}{x}\right)\left(- \frac{\left(x - 4\right)^{8} \sin^{4}{\left(x \right)}}{78125 x^{7} \left(x + 3\right)^{3} \sqrt{\left(4 x + 6\right)^{3}}} \right)$ Using some Trigonometric identities to simplify gives $LaTeX: y' = \left(\frac{4}{\tan{\left(x \right)}} + \frac{8}{x - 4}- \frac{6}{4 x + 6} - \frac{3}{x + 3} - \frac{7}{x}\right)\left(- \frac{\left(x - 4\right)^{8} \sin^{4}{\left(x \right)}}{78125 x^{7} \left(x + 3\right)^{3} \sqrt{\left(4 x + 6\right)^{3}}} \right)$