Find the derivative of LaTeX: \displaystyle y = \frac{\left(1 - 8 x\right)^{8} \left(4 x - 3\right)^{2}}{\sin^{8}{\left(x \right)} \cos^{3}{\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(1 - 8 x\right)^{8} \left(4 x - 3\right)^{2}}{\sin^{8}{\left(x \right)} \cos^{3}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 8 \ln{\left(1 - 8 x \right)} + 2 \ln{\left(4 x - 3 \right)}- 8 \ln{\left(\sin{\left(x \right)} \right)} - 3 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{3 \sin{\left(x \right)}}{\cos{\left(x \right)}} - \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{8}{4 x - 3} - \frac{64}{1 - 8 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{3 \sin{\left(x \right)}}{\cos{\left(x \right)}} - \frac{8 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{8}{4 x - 3} - \frac{64}{1 - 8 x}\right)\left(\frac{\left(1 - 8 x\right)^{8} \left(4 x - 3\right)^{2}}{\sin^{8}{\left(x \right)} \cos^{3}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{8}{4 x - 3} - \frac{64}{1 - 8 x}3 \tan{\left(x \right)} - \frac{8}{\tan{\left(x \right)}}\right)\left(\frac{\left(1 - 8 x\right)^{8} \left(4 x - 3\right)^{2}}{\sin^{8}{\left(x \right)} \cos^{3}{\left(x \right)}} \right)