Find the derivative of LaTeX: \displaystyle y = \frac{\left(9 x - 5\right)^{7} \sqrt{\left(5 x + 2\right)^{3}} e^{x}}{\left(x + 3\right)^{2} \left(4 x + 3\right)^{5} \cos^{6}{\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 - 5\right)^{7} \sqrt{\left(5 x + 2\right)^{3}} e^{x}}{\left(x + 3\right)^{2} \left(4 x + 3\right)^{5} \cos^{6}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + \frac{3 \ln{\left(5 x + 2 \right)}}{2} + 7 \ln{\left(9 x - 5 \right)}- 2 \ln{\left(x + 3 \right)} - 5 \ln{\left(4 x + 3 \right)} - 6 \ln{\left(\cos{\left(x \right)} \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)}} + 1 + \frac{63}{9 x - 5} + \frac{15}{2 \left(5 x + 2\right)} - \frac{20}{4 x + 3} - \frac{2}{x + 3} 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)}} + 1 + \frac{63}{9 x - 5} + \frac{15}{2 \left(5 x + 2\right)} - \frac{20}{4 x + 3} - \frac{2}{x + 3}\right)\left(\frac{\left(9 x - 5\right)^{7} \sqrt{\left(5 x + 2\right)^{3}} e^{x}}{\left(x + 3\right)^{2} \left(4 x + 3\right)^{5} \cos^{6}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{63}{9 x - 5} + \frac{15}{2 \left(5 x + 2\right)}6 \tan{\left(x \right)} - \frac{20}{4 x + 3} - \frac{2}{x + 3}\right)\left(\frac{\left(9 x - 5\right)^{7} \sqrt{\left(5 x + 2\right)^{3}} e^{x}}{\left(x + 3\right)^{2} \left(4 x + 3\right)^{5} \cos^{6}{\left(x \right)}} \right)