Find the derivative of LaTeX: \displaystyle y = \frac{\left(9 x + 9\right)^{8} \sqrt{\left(5 x + 4\right)^{3}}}{\left(- 8 x - 1\right)^{2} \left(x - 7\right)^{7} \cos^{7}{\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(9 x + 9\right)^{8} \sqrt{\left(5 x + 4\right)^{3}}}{\left(- 8 x - 1\right)^{2} \left(x - 7\right)^{7} \cos^{7}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{3 \ln{\left(5 x + 4 \right)}}{2} + 8 \ln{\left(9 x + 9 \right)}- 2 \ln{\left(- 8 x - 1 \right)} - 7 \ln{\left(x - 7 \right)} - 7 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{72}{9 x + 9} + \frac{15}{2 \left(5 x + 4\right)} - \frac{7}{x - 7} + \frac{16}{- 8 x - 1} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{72}{9 x + 9} + \frac{15}{2 \left(5 x + 4\right)} - \frac{7}{x - 7} + \frac{16}{- 8 x - 1}\right)\left(\frac{\left(9 x + 9\right)^{8} \sqrt{\left(5 x + 4\right)^{3}}}{\left(- 8 x - 1\right)^{2} \left(x - 7\right)^{7} \cos^{7}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{72}{9 x + 9} + \frac{15}{2 \left(5 x + 4\right)}7 \tan{\left(x \right)} - \frac{7}{x - 7} + \frac{16}{- 8 x - 1}\right)\left(\frac{\left(9 x + 9\right)^{8} \sqrt{\left(5 x + 4\right)^{3}}}{\left(- 8 x - 1\right)^{2} \left(x - 7\right)^{7} \cos^{7}{\left(x \right)}} \right)