Find the derivative of LaTeX: \displaystyle y = \frac{\left(4 x - 5\right)^{5} \left(7 x - 2\right)^{6} \sqrt{\left(7 x + 7\right)^{3}} e^{x}}{\sin^{6}{\left(x \right)} \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(4 x - 5\right)^{5} \left(7 x - 2\right)^{6} \sqrt{\left(7 x + 7\right)^{3}} e^{x}}{\sin^{6}{\left(x \right)} \cos^{4}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 5 \ln{\left(4 x - 5 \right)} + 6 \ln{\left(7 x - 2 \right)} + \frac{3 \ln{\left(7 x + 7 \right)}}{2}- 6 \ln{\left(\sin{\left(x \right)} \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)}} + 1 - \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{21}{2 \left(7 x + 7\right)} + \frac{42}{7 x - 2} + \frac{20}{4 x - 5} 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)}} + 1 - \frac{6 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{21}{2 \left(7 x + 7\right)} + \frac{42}{7 x - 2} + \frac{20}{4 x - 5}\right)\left(\frac{\left(4 x - 5\right)^{5} \left(7 x - 2\right)^{6} \sqrt{\left(7 x + 7\right)^{3}} e^{x}}{\sin^{6}{\left(x \right)} \cos^{4}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{21}{2 \left(7 x + 7\right)} + \frac{42}{7 x - 2} + \frac{20}{4 x - 5}4 \tan{\left(x \right)} - \frac{6}{\tan{\left(x \right)}}\right)\left(\frac{\left(4 x - 5\right)^{5} \left(7 x - 2\right)^{6} \sqrt{\left(7 x + 7\right)^{3}} e^{x}}{\sin^{6}{\left(x \right)} \cos^{4}{\left(x \right)}} \right)