Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 6 x - 2\right)^{8} \sqrt{3 x + 7} e^{x} \sin^{8}{\left(x \right)}}{\left(x - 4\right)^{5} \left(7 x - 7\right)^{4} \cos^{4}{\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(- 6 x - 2\right)^{8} \sqrt{3 x + 7} e^{x} \sin^{8}{\left(x \right)}}{\left(x - 4\right)^{5} \left(7 x - 7\right)^{4} \cos^{4}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 8 \ln{\left(- 6 x - 2 \right)} + \frac{\ln{\left(3 x + 7 \right)}}{2} + 8 \ln{\left(\sin{\left(x \right)} \right)}- 5 \ln{\left(x - 4 \right)} - 4 \ln{\left(7 x - 7 \right)} - 4 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{28}{7 x - 7} + \frac{3}{2 \left(3 x + 7\right)} - \frac{5}{x - 4} - \frac{48}{- 6 x - 2} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{28}{7 x - 7} + \frac{3}{2 \left(3 x + 7\right)} - \frac{5}{x - 4} - \frac{48}{- 6 x - 2}\right)\left(\frac{\left(- 6 x - 2\right)^{8} \sqrt{3 x + 7} e^{x} \sin^{8}{\left(x \right)}}{\left(x - 4\right)^{5} \left(7 x - 7\right)^{4} \cos^{4}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{8}{\tan{\left(x \right)}} + \frac{3}{2 \left(3 x + 7\right)} - \frac{48}{- 6 x - 2}4 \tan{\left(x \right)} - \frac{28}{7 x - 7} - \frac{5}{x - 4}\right)\left(\frac{\left(- 6 x - 2\right)^{8} \sqrt{3 x + 7} e^{x} \sin^{8}{\left(x \right)}}{\left(x - 4\right)^{5} \left(7 x - 7\right)^{4} \cos^{4}{\left(x \right)}} \right)