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