Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 9\right)^{7} \sin^{6}{\left(x \right)}}{\left(4 x + 7\right)^{4} \sqrt{\left(2 x + 8\right)^{3}} \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(x - 9\right)^{7} \sin^{6}{\left(x \right)}}{\left(4 x + 7\right)^{4} \sqrt{\left(2 x + 8\right)^{3}} \cos^{3}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 7 \ln{\left(x - 9 \right)} + 6 \ln{\left(\sin{\left(x \right)} \right)}- \frac{3 \ln{\left(2 x + 8 \right)}}{2} - 4 \ln{\left(4 x + 7 \right)} - 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{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{16}{4 x + 7} - \frac{3}{2 x + 8} + \frac{7}{x - 9} 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{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{16}{4 x + 7} - \frac{3}{2 x + 8} + \frac{7}{x - 9}\right)\left(\frac{\left(x - 9\right)^{7} \sin^{6}{\left(x \right)}}{\left(4 x + 7\right)^{4} \sqrt{\left(2 x + 8\right)^{3}} \cos^{3}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{6}{\tan{\left(x \right)}} + \frac{7}{x - 9}3 \tan{\left(x \right)} - \frac{16}{4 x + 7} - \frac{3}{2 x + 8}\right)\left(\frac{\left(x - 9\right)^{7} \sin^{6}{\left(x \right)}}{\left(4 x + 7\right)^{4} \sqrt{\left(2 x + 8\right)^{3}} \cos^{3}{\left(x \right)}} \right)