Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 3\right)^{6} \sqrt{5 x + 6} \cos^{6}{\left(x \right)}}{\left(6 x - 4\right)^{7} \sin^{8}{\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)^{6} \sqrt{5 x + 6} \cos^{6}{\left(x \right)}}{\left(6 x - 4\right)^{7} \sin^{8}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 6 \ln{\left(x + 3 \right)} + \frac{\ln{\left(5 x + 6 \right)}}{2} + 6 \ln{\left(\cos{\left(x \right)} \right)}- 7 \ln{\left(6 x - 4 \right)} - 8 \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{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{42}{6 x - 4} + \frac{5}{2 \left(5 x + 6\right)} + \frac{6}{x + 3} 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{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{42}{6 x - 4} + \frac{5}{2 \left(5 x + 6\right)} + \frac{6}{x + 3}\right)\left(\frac{\left(x + 3\right)^{6} \sqrt{5 x + 6} \cos^{6}{\left(x \right)}}{\left(6 x - 4\right)^{7} \sin^{8}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 6 \tan{\left(x \right)} + \frac{5}{2 \left(5 x + 6\right)} + \frac{6}{x + 3}- \frac{8}{\tan{\left(x \right)}} - \frac{42}{6 x - 4}\right)\left(\frac{\left(x + 3\right)^{6} \sqrt{5 x + 6} \cos^{6}{\left(x \right)}}{\left(6 x - 4\right)^{7} \sin^{8}{\left(x \right)}} \right)