Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 3\right)^{4} \sqrt{\left(x + 8\right)^{7}} \cos^{4}{\left(x \right)}}{\left(7 - 4 x\right)^{8} \sin^{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(x - 3\right)^{4} \sqrt{\left(x + 8\right)^{7}} \cos^{4}{\left(x \right)}}{\left(7 - 4 x\right)^{8} \sin^{4}{\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(x + 8 \right)}}{2} + 4 \ln{\left(\cos{\left(x \right)} \right)}- 8 \ln{\left(7 - 4 x \right)} - 4 \ln{\left(\sin{\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{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{7}{2 \left(x + 8\right)} + \frac{4}{x - 3} + \frac{32}{7 - 4 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{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{7}{2 \left(x + 8\right)} + \frac{4}{x - 3} + \frac{32}{7 - 4 x}\right)\left(\frac{\left(x - 3\right)^{4} \sqrt{\left(x + 8\right)^{7}} \cos^{4}{\left(x \right)}}{\left(7 - 4 x\right)^{8} \sin^{4}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 4 \tan{\left(x \right)} + \frac{7}{2 \left(x + 8\right)} + \frac{4}{x - 3}- \frac{4}{\tan{\left(x \right)}} + \frac{32}{7 - 4 x}\right)\left(\frac{\left(x - 3\right)^{4} \sqrt{\left(x + 8\right)^{7}} \cos^{4}{\left(x \right)}}{\left(7 - 4 x\right)^{8} \sin^{4}{\left(x \right)}} \right)