Find the derivative of LaTeX: \displaystyle y = \frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\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{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 7 \ln{\left(\sin{\left(x \right)} \right)} + 4 \ln{\left(\cos{\left(x \right)} \right)}- x - 6 \ln{\left(2 - 8 x \right)} - 5 \ln{\left(3 x + 8 \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{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{3 x + 8} + \frac{48}{2 - 8 x} 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{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{3 x + 8} + \frac{48}{2 - 8 x}\right)\left(\frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 4 \tan{\left(x \right)} + \frac{7}{\tan{\left(x \right)}}-1 - \frac{15}{3 x + 8} + \frac{48}{2 - 8 x}\right)\left(\frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} \right)