Find the derivative of LaTeX: \displaystyle y = \frac{\left(x + 1\right)^{3} \left(5 x - 4\right)^{6} \sqrt{\left(8 x + 7\right)^{3}}}{\sin^{5}{\left(x \right)} \cos^{7}{\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 + 1\right)^{3} \left(5 x - 4\right)^{6} \sqrt{\left(8 x + 7\right)^{3}}}{\sin^{5}{\left(x \right)} \cos^{7}{\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 + 1 \right)} + 6 \ln{\left(5 x - 4 \right)} + \frac{3 \ln{\left(8 x + 7 \right)}}{2}- 5 \ln{\left(\sin{\left(x \right)} \right)} - 7 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{12}{8 x + 7} + \frac{30}{5 x - 4} + \frac{3}{x + 1} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{7 \sin{\left(x \right)}}{\cos{\left(x \right)}} - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{12}{8 x + 7} + \frac{30}{5 x - 4} + \frac{3}{x + 1}\right)\left(\frac{\left(x + 1\right)^{3} \left(5 x - 4\right)^{6} \sqrt{\left(8 x + 7\right)^{3}}}{\sin^{5}{\left(x \right)} \cos^{7}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{12}{8 x + 7} + \frac{30}{5 x - 4} + \frac{3}{x + 1}7 \tan{\left(x \right)} - \frac{5}{\tan{\left(x \right)}}\right)\left(\frac{\left(x + 1\right)^{3} \left(5 x - 4\right)^{6} \sqrt{\left(8 x + 7\right)^{3}}}{\sin^{5}{\left(x \right)} \cos^{7}{\left(x \right)}} \right)