Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 4 x - 5\right)^{4} e^{x} \sin^{6}{\left(x \right)}}{\left(7 x + 2\right)^{4} \sqrt{\left(8 x + 5\right)^{5}} \cos^{7}{\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(- 4 x - 5\right)^{4} e^{x} \sin^{6}{\left(x \right)}}{\left(7 x + 2\right)^{4} \sqrt{\left(8 x + 5\right)^{5}} \cos^{7}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 4 \ln{\left(- 4 x - 5 \right)} + 6 \ln{\left(\sin{\left(x \right)} \right)}- 4 \ln{\left(7 x + 2 \right)} - \frac{5 \ln{\left(8 x + 5 \right)}}{2} - 7 \ln{\left(\cos{\left(x \right)} \right)} 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{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{20}{8 x + 5} - \frac{28}{7 x + 2} - \frac{16}{- 4 x - 5} 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{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{20}{8 x + 5} - \frac{28}{7 x + 2} - \frac{16}{- 4 x - 5}\right)\left(\frac{\left(- 4 x - 5\right)^{4} e^{x} \sin^{6}{\left(x \right)}}{\left(7 x + 2\right)^{4} \sqrt{\left(8 x + 5\right)^{5}} \cos^{7}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{6}{\tan{\left(x \right)}} - \frac{16}{- 4 x - 5}7 \tan{\left(x \right)} - \frac{20}{8 x + 5} - \frac{28}{7 x + 2}\right)\left(\frac{\left(- 4 x - 5\right)^{4} e^{x} \sin^{6}{\left(x \right)}}{\left(7 x + 2\right)^{4} \sqrt{\left(8 x + 5\right)^{5}} \cos^{7}{\left(x \right)}} \right)