Find the derivative of LaTeX: \displaystyle y = \frac{\left(- 8 x - 6\right)^{3} e^{x} \sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)}}{\left(- 8 x - 2\right)^{6} \sqrt{\left(6 x + 8\right)^{3}}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(- 8 x - 6\right)^{3} e^{x} \sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)}}{\left(- 8 x - 2\right)^{6} \sqrt{\left(6 x + 8\right)^{3}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 3 \ln{\left(- 8 x - 6 \right)} + 2 \ln{\left(\sin{\left(x \right)} \right)} + 5 \ln{\left(\cos{\left(x \right)} \right)}- 6 \ln{\left(- 8 x - 2 \right)} - \frac{3 \ln{\left(6 x + 8 \right)}}{2} 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{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{9}{6 x + 8} + \frac{48}{- 8 x - 2} - \frac{24}{- 8 x - 6} 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{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{9}{6 x + 8} + \frac{48}{- 8 x - 2} - \frac{24}{- 8 x - 6}\right)\left(\frac{\left(- 8 x - 6\right)^{3} e^{x} \sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)}}{\left(- 8 x - 2\right)^{6} \sqrt{\left(6 x + 8\right)^{3}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 5 \tan{\left(x \right)} + 1 + \frac{2}{\tan{\left(x \right)}} - \frac{24}{- 8 x - 6}- \frac{9}{6 x + 8} + \frac{48}{- 8 x - 2}\right)\left(\frac{\left(- 8 x - 6\right)^{3} e^{x} \sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)}}{\left(- 8 x - 2\right)^{6} \sqrt{\left(6 x + 8\right)^{3}}} \right)