Find the derivative of LaTeX: \displaystyle y = \frac{\left(6 - 8 x\right)^{8} e^{x} \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(7 - 8 x\right)^{8} \sqrt{\left(7 x + 3\right)^{7}}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(6 - 8 x\right)^{8} e^{x} \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(7 - 8 x\right)^{8} \sqrt{\left(7 x + 3\right)^{7}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 8 \ln{\left(6 - 8 x \right)} + 6 \ln{\left(\sin{\left(x \right)} \right)} + 2 \ln{\left(\cos{\left(x \right)} \right)}- 8 \ln{\left(7 - 8 x \right)} - \frac{7 \ln{\left(7 x + 3 \right)}}{2} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{49}{2 \left(7 x + 3\right)} + \frac{64}{7 - 8 x} - \frac{64}{6 - 8 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} + 1 + \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{49}{2 \left(7 x + 3\right)} + \frac{64}{7 - 8 x} - \frac{64}{6 - 8 x}\right)\left(\frac{\left(6 - 8 x\right)^{8} e^{x} \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(7 - 8 x\right)^{8} \sqrt{\left(7 x + 3\right)^{7}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 2 \tan{\left(x \right)} + 1 + \frac{6}{\tan{\left(x \right)}} - \frac{64}{6 - 8 x}- \frac{49}{2 \left(7 x + 3\right)} + \frac{64}{7 - 8 x}\right)\left(\frac{\left(6 - 8 x\right)^{8} e^{x} \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}}{\left(7 - 8 x\right)^{8} \sqrt{\left(7 x + 3\right)^{7}}} \right)