Find the derivative of LaTeX: \displaystyle y = \frac{\left(3 x - 8\right)^{5} \sin^{3}{\left(x \right)}}{\left(x - 3\right)^{6} \sqrt{\left(6 x + 5\right)^{5}} \cos^{3}{\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(3 x - 8\right)^{5} \sin^{3}{\left(x \right)}}{\left(x - 3\right)^{6} \sqrt{\left(6 x + 5\right)^{5}} \cos^{3}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 5 \ln{\left(3 x - 8 \right)} + 3 \ln{\left(\sin{\left(x \right)} \right)}- 6 \ln{\left(x - 3 \right)} - \frac{5 \ln{\left(6 x + 5 \right)}}{2} - 3 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{3 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{6 x + 5} + \frac{15}{3 x - 8} - \frac{6}{x - 3} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{3 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{3 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{6 x + 5} + \frac{15}{3 x - 8} - \frac{6}{x - 3}\right)\left(\frac{\left(3 x - 8\right)^{5} \sin^{3}{\left(x \right)}}{\left(x - 3\right)^{6} \sqrt{\left(6 x + 5\right)^{5}} \cos^{3}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{3}{\tan{\left(x \right)}} + \frac{15}{3 x - 8}3 \tan{\left(x \right)} - \frac{15}{6 x + 5} - \frac{6}{x - 3}\right)\left(\frac{\left(3 x - 8\right)^{5} \sin^{3}{\left(x \right)}}{\left(x - 3\right)^{6} \sqrt{\left(6 x + 5\right)^{5}} \cos^{3}{\left(x \right)}} \right)