Find the derivative of LaTeX: \displaystyle y = \frac{46656 x^{6} \sqrt{5 x + 7} \left(6 x - 4\right)^{3} e^{x}}{\sin^{5}{\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{46656 x^{6} \sqrt{5 x + 7} \left(6 x - 4\right)^{3} e^{x}}{\sin^{5}{\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) = x + 6 \ln{\left(x \right)} + \frac{\ln{\left(5 x + 7 \right)}}{2} + 3 \ln{\left(6 x - 4 \right)} + 6 \ln{\left(6 \right)}- 5 \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)}} + 1 - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{18}{6 x - 4} + \frac{5}{2 \left(5 x + 7\right)} + \frac{6}{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)}} + 1 - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{18}{6 x - 4} + \frac{5}{2 \left(5 x + 7\right)} + \frac{6}{x}\right)\left(\frac{46656 x^{6} \sqrt{5 x + 7} \left(6 x - 4\right)^{3} e^{x}}{\sin^{5}{\left(x \right)} \cos^{3}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{18}{6 x - 4} + \frac{5}{2 \left(5 x + 7\right)} + \frac{6}{x}3 \tan{\left(x \right)} - \frac{5}{\tan{\left(x \right)}}\right)\left(\frac{46656 x^{6} \sqrt{5 x + 7} \left(6 x - 4\right)^{3} e^{x}}{\sin^{5}{\left(x \right)} \cos^{3}{\left(x \right)}} \right)