Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 2\right)^{6} \sin^{5}{\left(x \right)} \cos^{6}{\left(x \right)}}{4096 x^{4} \sqrt{\left(9 x + 5\right)^{7}}}

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 - 2\right)^{6} \sin^{5}{\left(x \right)} \cos^{6}{\left(x \right)}}{4096 x^{4} \sqrt{\left(9 x + 5\right)^{7}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 6 \ln{\left(x - 2 \right)} + 5 \ln{\left(\sin{\left(x \right)} \right)} + 6 \ln{\left(\cos{\left(x \right)} \right)}- 4 \ln{\left(x \right)} - \frac{7 \ln{\left(9 x + 5 \right)}}{2} - 12 \ln{\left(2 \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{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{63}{2 \left(9 x + 5\right)} + \frac{6}{x - 2} - \frac{4}{x} 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{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{63}{2 \left(9 x + 5\right)} + \frac{6}{x - 2} - \frac{4}{x}\right)\left(\frac{\left(x - 2\right)^{6} \sin^{5}{\left(x \right)} \cos^{6}{\left(x \right)}}{4096 x^{4} \sqrt{\left(9 x + 5\right)^{7}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 6 \tan{\left(x \right)} + \frac{5}{\tan{\left(x \right)}} + \frac{6}{x - 2}- \frac{63}{2 \left(9 x + 5\right)} - \frac{4}{x}\right)\left(\frac{\left(x - 2\right)^{6} \sin^{5}{\left(x \right)} \cos^{6}{\left(x \right)}}{4096 x^{4} \sqrt{\left(9 x + 5\right)^{7}}} \right)