Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 3\right)^{3} \sin^{7}{\left(x \right)}}{\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 + 3\right)^{3} \sin^{7}{\left(x \right)}}{\cos^{8}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 3 \ln{\left(x + 3 \right)} + 7 \ln{\left(\sin{\left(x \right)} \right)}- 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{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{3}{x + 3} 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{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{3}{x + 3}\right)\left(\frac{\left(x + 3\right)^{3} \sin^{7}{\left(x \right)}}{\cos^{8}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{7}{\tan{\left(x \right)}} + \frac{3}{x + 3}8 \tan{\left(x \right)}\right)\left(\frac{\left(x + 3\right)^{3} \sin^{7}{\left(x \right)}}{\cos^{8}{\left(x \right)}} \right)