Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 1\right)^{5} \sin^{2}{\left(x \right)}}{\left(2 x - 2\right)^{8} \sqrt{\left(5 x + 9\right)^{7}} \cos^{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 + 1\right)^{5} \sin^{2}{\left(x \right)}}{\left(2 x - 2\right)^{8} \sqrt{\left(5 x + 9\right)^{7}} \cos^{8}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 5 \ln{\left(x + 1 \right)} + 2 \ln{\left(\sin{\left(x \right)} \right)}- 8 \ln{\left(2 x - 2 \right)} - \frac{7 \ln{\left(5 x + 9 \right)}}{2} - 8 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{8 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{35}{2 \left(5 x + 9\right)} - \frac{16}{2 x - 2} + \frac{5}{x + 1} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{8 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{35}{2 \left(5 x + 9\right)} - \frac{16}{2 x - 2} + \frac{5}{x + 1}\right)\left(\frac{\left(x + 1\right)^{5} \sin^{2}{\left(x \right)}}{\left(2 x - 2\right)^{8} \sqrt{\left(5 x + 9\right)^{7}} \cos^{8}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{2}{\tan{\left(x \right)}} + \frac{5}{x + 1}8 \tan{\left(x \right)} - \frac{35}{2 \left(5 x + 9\right)} - \frac{16}{2 x - 2}\right)\left(\frac{\left(x + 1\right)^{5} \sin^{2}{\left(x \right)}}{\left(2 x - 2\right)^{8} \sqrt{\left(5 x + 9\right)^{7}} \cos^{8}{\left(x \right)}} \right)