Find the derivative of LaTeX: \displaystyle y = \frac{\sqrt{\left(7 x + 1\right)^{7}} \cos^{3}{\left(x \right)}}{\left(x + 5\right)^{3}}

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(7 x + 1\right)^{7}} \cos^{3}{\left(x \right)}}{\left(x + 5\right)^{3}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{7 \ln{\left(7 x + 1 \right)}}{2} + 3 \ln{\left(\cos{\left(x \right)} \right)}- 3 \ln{\left(x + 5 \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{49}{2 \left(7 x + 1\right)} - \frac{3}{x + 5} 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{49}{2 \left(7 x + 1\right)} - \frac{3}{x + 5}\right)\left(\frac{\sqrt{\left(7 x + 1\right)^{7}} \cos^{3}{\left(x \right)}}{\left(x + 5\right)^{3}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 3 \tan{\left(x \right)} + \frac{49}{2 \left(7 x + 1\right)}- \frac{3}{x + 5}\right)\left(\frac{\sqrt{\left(7 x + 1\right)^{7}} \cos^{3}{\left(x \right)}}{\left(x + 5\right)^{3}} \right)