Find the derivative of LaTeX: \displaystyle y = \frac{\sqrt{\left(6 x + 1\right)^{7}} \cos^{5}{\left(x \right)}}{\left(x + 4\right)^{3} \left(4 x - 4\right)^{2} \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{\sqrt{\left(6 x + 1\right)^{7}} \cos^{5}{\left(x \right)}}{\left(x + 4\right)^{3} \left(4 x - 4\right)^{2} \sin^{4}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{7 \ln{\left(6 x + 1 \right)}}{2} + 5 \ln{\left(\cos{\left(x \right)} \right)}- 3 \ln{\left(x + 4 \right)} - 2 \ln{\left(4 x - 4 \right)} - 4 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} - \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{21}{6 x + 1} - \frac{8}{4 x - 4} - \frac{3}{x + 4} 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)}} - \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{21}{6 x + 1} - \frac{8}{4 x - 4} - \frac{3}{x + 4}\right)\left(\frac{\sqrt{\left(6 x + 1\right)^{7}} \cos^{5}{\left(x \right)}}{\left(x + 4\right)^{3} \left(4 x - 4\right)^{2} \sin^{4}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 5 \tan{\left(x \right)} + \frac{21}{6 x + 1}- \frac{4}{\tan{\left(x \right)}} - \frac{8}{4 x - 4} - \frac{3}{x + 4}\right)\left(\frac{\sqrt{\left(6 x + 1\right)^{7}} \cos^{5}{\left(x \right)}}{\left(x + 4\right)^{3} \left(4 x - 4\right)^{2} \sin^{4}{\left(x \right)}} \right)