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