Find the derivative of LaTeX: \displaystyle y = \frac{\left(3 x - 4\right)^{4} \sqrt{\left(9 x + 1\right)^{3}} e^{- x}}{\left(4 x - 7\right)^{8} \sin^{4}{\left(x \right)} \cos^{5}{\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(3 x - 4\right)^{4} \sqrt{\left(9 x + 1\right)^{3}} e^{- x}}{\left(4 x - 7\right)^{8} \sin^{4}{\left(x \right)} \cos^{5}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(3 x - 4 \right)} + \frac{3 \ln{\left(9 x + 1 \right)}}{2}- x - 8 \ln{\left(4 x - 7 \right)} - 4 \ln{\left(\sin{\left(x \right)} \right)} - 5 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 - \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{27}{2 \left(9 x + 1\right)} - \frac{32}{4 x - 7} + \frac{12}{3 x - 4} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 - \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{27}{2 \left(9 x + 1\right)} - \frac{32}{4 x - 7} + \frac{12}{3 x - 4}\right)\left(\frac{\left(3 x - 4\right)^{4} \sqrt{\left(9 x + 1\right)^{3}} e^{- x}}{\left(4 x - 7\right)^{8} \sin^{4}{\left(x \right)} \cos^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{27}{2 \left(9 x + 1\right)} + \frac{12}{3 x - 4}5 \tan{\left(x \right)} - 1 - \frac{4}{\tan{\left(x \right)}} - \frac{32}{4 x - 7}\right)\left(\frac{\left(3 x - 4\right)^{4} \sqrt{\left(9 x + 1\right)^{3}} e^{- x}}{\left(4 x - 7\right)^{8} \sin^{4}{\left(x \right)} \cos^{5}{\left(x \right)}} \right)