Find the derivative of LaTeX: \displaystyle y = \frac{\left(7 x - 2\right)^{4} e^{- x} \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(- x - 9\right)^{8} \left(x - 8\right)^{2} \sqrt{\left(4 x + 3\right)^{5}}}

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 - 2\right)^{4} e^{- x} \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(- x - 9\right)^{8} \left(x - 8\right)^{2} \sqrt{\left(4 x + 3\right)^{5}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(7 x - 2 \right)} + 6 \ln{\left(\sin{\left(x \right)} \right)} + 2 \ln{\left(\cos{\left(x \right)} \right)}- x - 8 \ln{\left(- x - 9 \right)} - 2 \ln{\left(x - 8 \right)} - \frac{5 \ln{\left(4 x + 3 \right)}}{2} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{28}{7 x - 2} - \frac{10}{4 x + 3} - \frac{2}{x - 8} + \frac{8}{- x - 9} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{28}{7 x - 2} - \frac{10}{4 x + 3} - \frac{2}{x - 8} + \frac{8}{- x - 9}\right)\left(\frac{\left(7 x - 2\right)^{4} e^{- x} \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(- x - 9\right)^{8} \left(x - 8\right)^{2} \sqrt{\left(4 x + 3\right)^{5}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 2 \tan{\left(x \right)} + \frac{6}{\tan{\left(x \right)}} + \frac{28}{7 x - 2}-1 - \frac{10}{4 x + 3} - \frac{2}{x - 8} + \frac{8}{- x - 9}\right)\left(\frac{\left(7 x - 2\right)^{4} e^{- x} \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(- x - 9\right)^{8} \left(x - 8\right)^{2} \sqrt{\left(4 x + 3\right)^{5}}} \right)