Find the derivative of LaTeX: \displaystyle y = \frac{\left(7 x + 8\right)^{2} \sqrt{\left(5 x + 7\right)^{5}} \sin^{2}{\left(x \right)}}{\left(8 - 7 x\right)^{2} \left(x + 3\right)^{7} \cos^{4}{\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(7 x + 8\right)^{2} \sqrt{\left(5 x + 7\right)^{5}} \sin^{2}{\left(x \right)}}{\left(8 - 7 x\right)^{2} \left(x + 3\right)^{7} \cos^{4}{\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(5 x + 7 \right)}}{2} + 2 \ln{\left(7 x + 8 \right)} + 2 \ln{\left(\sin{\left(x \right)} \right)}- 2 \ln{\left(8 - 7 x \right)} - 7 \ln{\left(x + 3 \right)} - 4 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{14}{7 x + 8} + \frac{25}{2 \left(5 x + 7\right)} - \frac{7}{x + 3} + \frac{14}{8 - 7 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{2 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{14}{7 x + 8} + \frac{25}{2 \left(5 x + 7\right)} - \frac{7}{x + 3} + \frac{14}{8 - 7 x}\right)\left(\frac{\left(7 x + 8\right)^{2} \sqrt{\left(5 x + 7\right)^{5}} \sin^{2}{\left(x \right)}}{\left(8 - 7 x\right)^{2} \left(x + 3\right)^{7} \cos^{4}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{2}{\tan{\left(x \right)}} + \frac{14}{7 x + 8} + \frac{25}{2 \left(5 x + 7\right)}4 \tan{\left(x \right)} - \frac{7}{x + 3} + \frac{14}{8 - 7 x}\right)\left(\frac{\left(7 x + 8\right)^{2} \sqrt{\left(5 x + 7\right)^{5}} \sin^{2}{\left(x \right)}}{\left(8 - 7 x\right)^{2} \left(x + 3\right)^{7} \cos^{4}{\left(x \right)}} \right)