Find the derivative of LaTeX: \displaystyle y = \frac{\left(7 x + 3\right)^{6} \left(8 x + 5\right)^{3} \sqrt{\left(6 x + 6\right)^{5}} e^{x}}{\left(2 - x\right)^{4} \left(x + 5\right)^{6}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(7 x + 3\right)^{6} \left(8 x + 5\right)^{3} \sqrt{\left(6 x + 6\right)^{5}} e^{x}}{\left(2 - x\right)^{4} \left(x + 5\right)^{6}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + \frac{5 \ln{\left(6 x + 6 \right)}}{2} + 6 \ln{\left(7 x + 3 \right)} + 3 \ln{\left(8 x + 5 \right)}- 4 \ln{\left(2 - x \right)} - 6 \ln{\left(x + 5 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = 1 + \frac{24}{8 x + 5} + \frac{42}{7 x + 3} + \frac{15}{6 x + 6} - \frac{6}{x + 5} + \frac{4}{2 - x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(1 + \frac{24}{8 x + 5} + \frac{42}{7 x + 3} + \frac{15}{6 x + 6} - \frac{6}{x + 5} + \frac{4}{2 - x}\right)\left(\frac{\left(7 x + 3\right)^{6} \left(8 x + 5\right)^{3} \sqrt{\left(6 x + 6\right)^{5}} e^{x}}{\left(2 - x\right)^{4} \left(x + 5\right)^{6}} \right)