Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 3\right)^{4} \sqrt{\left(3 x + 1\right)^{7}} \sin^{5}{\left(x \right)}}{\left(x + 5\right)^{8} \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(x + 3\right)^{4} \sqrt{\left(3 x + 1\right)^{7}} \sin^{5}{\left(x \right)}}{\left(x + 5\right)^{8} \cos^{6}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(x + 3 \right)} + \frac{7 \ln{\left(3 x + 1 \right)}}{2} + 5 \ln{\left(\sin{\left(x \right)} \right)}- 8 \ln{\left(x + 5 \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)}} + \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{21}{2 \left(3 x + 1\right)} - \frac{8}{x + 5} + \frac{4}{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)}} + \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{21}{2 \left(3 x + 1\right)} - \frac{8}{x + 5} + \frac{4}{x + 3}\right)\left(\frac{\left(x + 3\right)^{4} \sqrt{\left(3 x + 1\right)^{7}} \sin^{5}{\left(x \right)}}{\left(x + 5\right)^{8} \cos^{6}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{5}{\tan{\left(x \right)}} + \frac{21}{2 \left(3 x + 1\right)} + \frac{4}{x + 3}6 \tan{\left(x \right)} - \frac{8}{x + 5}\right)\left(\frac{\left(x + 3\right)^{4} \sqrt{\left(3 x + 1\right)^{7}} \sin^{5}{\left(x \right)}}{\left(x + 5\right)^{8} \cos^{6}{\left(x \right)}} \right)