Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 7 x - 3\right)^{8} \left(- 2 x - 1\right)^{8} \left(4 x - 1\right)^{7} e^{x}}{\sin^{7}{\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(- 7 x - 3\right)^{8} \left(- 2 x - 1\right)^{8} \left(4 x - 1\right)^{7} e^{x}}{\sin^{7}{\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) = x + 8 \ln{\left(- 7 x - 3 \right)} + 8 \ln{\left(- 2 x - 1 \right)} + 7 \ln{\left(4 x - 1 \right)}- 7 \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{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{28}{4 x - 1} - \frac{16}{- 2 x - 1} - \frac{56}{- 7 x - 3} 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{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{28}{4 x - 1} - \frac{16}{- 2 x - 1} - \frac{56}{- 7 x - 3}\right)\left(\frac{\left(- 7 x - 3\right)^{8} \left(- 2 x - 1\right)^{8} \left(4 x - 1\right)^{7} e^{x}}{\sin^{7}{\left(x \right)} \cos^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{28}{4 x - 1} - \frac{16}{- 2 x - 1} - \frac{56}{- 7 x - 3}5 \tan{\left(x \right)} - \frac{7}{\tan{\left(x \right)}}\right)\left(\frac{\left(- 7 x - 3\right)^{8} \left(- 2 x - 1\right)^{8} \left(4 x - 1\right)^{7} e^{x}}{\sin^{7}{\left(x \right)} \cos^{5}{\left(x \right)}} \right)