Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 4\right)^{2} \left(7 x + 1\right)^{4} e^{- x} \sin^{7}{\left(x \right)}}{\left(1 - 3 x\right)^{8} \left(6 x - 4\right)^{2} \sqrt{\left(4 x + 3\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(x + 4\right)^{2} \left(7 x + 1\right)^{4} e^{- x} \sin^{7}{\left(x \right)}}{\left(1 - 3 x\right)^{8} \left(6 x - 4\right)^{2} \sqrt{\left(4 x + 3\right)^{7}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 2 \ln{\left(x + 4 \right)} + 4 \ln{\left(7 x + 1 \right)} + 7 \ln{\left(\sin{\left(x \right)} \right)}- x - 8 \ln{\left(1 - 3 x \right)} - \frac{7 \ln{\left(4 x + 3 \right)}}{2} - 2 \ln{\left(6 x - 4 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = -1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{28}{7 x + 1} - \frac{12}{6 x - 4} - \frac{14}{4 x + 3} + \frac{2}{x + 4} + \frac{24}{1 - 3 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(-1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{28}{7 x + 1} - \frac{12}{6 x - 4} - \frac{14}{4 x + 3} + \frac{2}{x + 4} + \frac{24}{1 - 3 x}\right)\left(\frac{\left(x + 4\right)^{2} \left(7 x + 1\right)^{4} e^{- x} \sin^{7}{\left(x \right)}}{\left(1 - 3 x\right)^{8} \left(6 x - 4\right)^{2} \sqrt{\left(4 x + 3\right)^{7}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{7}{\tan{\left(x \right)}} + \frac{28}{7 x + 1} + \frac{2}{x + 4}-1 - \frac{12}{6 x - 4} - \frac{14}{4 x + 3} + \frac{24}{1 - 3 x}\right)\left(\frac{\left(x + 4\right)^{2} \left(7 x + 1\right)^{4} e^{- x} \sin^{7}{\left(x \right)}}{\left(1 - 3 x\right)^{8} \left(6 x - 4\right)^{2} \sqrt{\left(4 x + 3\right)^{7}}} \right)