Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 3\right)^{8} \cos^{6}{\left(x \right)}}{\left(- 7 x - 6\right)^{6} \left(7 x - 9\right)^{7} \sin^{4}{\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(x + 3\right)^{8} \cos^{6}{\left(x \right)}}{\left(- 7 x - 6\right)^{6} \left(7 x - 9\right)^{7} \sin^{4}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 8 \ln{\left(x + 3 \right)} + 6 \ln{\left(\cos{\left(x \right)} \right)}- 6 \ln{\left(- 7 x - 6 \right)} - 7 \ln{\left(7 x - 9 \right)} - 4 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{6 \sin{\left(x \right)}}{\cos{\left(x \right)}} - \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{49}{7 x - 9} + \frac{8}{x + 3} + \frac{42}{- 7 x - 6} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{6 \sin{\left(x \right)}}{\cos{\left(x \right)}} - \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{49}{7 x - 9} + \frac{8}{x + 3} + \frac{42}{- 7 x - 6}\right)\left(\frac{\left(x + 3\right)^{8} \cos^{6}{\left(x \right)}}{\left(- 7 x - 6\right)^{6} \left(7 x - 9\right)^{7} \sin^{4}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 6 \tan{\left(x \right)} + \frac{8}{x + 3}- \frac{4}{\tan{\left(x \right)}} - \frac{49}{7 x - 9} + \frac{42}{- 7 x - 6}\right)\left(\frac{\left(x + 3\right)^{8} \cos^{6}{\left(x \right)}}{\left(- 7 x - 6\right)^{6} \left(7 x - 9\right)^{7} \sin^{4}{\left(x \right)}} \right)