Find the derivative of LaTeX: \displaystyle y = \frac{\left(5 - 3 x\right)^{2} \left(6 x - 4\right)^{5} e^{- x} \cos^{2}{\left(x \right)}}{\left(7 - 7 x\right)^{4} \sqrt{\left(6 x + 7\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(5 - 3 x\right)^{2} \left(6 x - 4\right)^{5} e^{- x} \cos^{2}{\left(x \right)}}{\left(7 - 7 x\right)^{4} \sqrt{\left(6 x + 7\right)^{7}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 2 \ln{\left(5 - 3 x \right)} + 5 \ln{\left(6 x - 4 \right)} + 2 \ln{\left(\cos{\left(x \right)} \right)}- x - 4 \ln{\left(7 - 7 x \right)} - \frac{7 \ln{\left(6 x + 7 \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{21}{6 x + 7} + \frac{30}{6 x - 4} + \frac{28}{7 - 7 x} - \frac{6}{5 - 3 x} 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{21}{6 x + 7} + \frac{30}{6 x - 4} + \frac{28}{7 - 7 x} - \frac{6}{5 - 3 x}\right)\left(\frac{\left(5 - 3 x\right)^{2} \left(6 x - 4\right)^{5} e^{- x} \cos^{2}{\left(x \right)}}{\left(7 - 7 x\right)^{4} \sqrt{\left(6 x + 7\right)^{7}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 2 \tan{\left(x \right)} + \frac{30}{6 x - 4} - \frac{6}{5 - 3 x}-1 - \frac{21}{6 x + 7} + \frac{28}{7 - 7 x}\right)\left(\frac{\left(5 - 3 x\right)^{2} \left(6 x - 4\right)^{5} e^{- x} \cos^{2}{\left(x \right)}}{\left(7 - 7 x\right)^{4} \sqrt{\left(6 x + 7\right)^{7}}} \right)