Find the derivative of LaTeX: \displaystyle y = \frac{\left(4 x + 3\right)^{6} e^{x} \sin^{4}{\left(x \right)}}{\left(- 6 x - 4\right)^{2} \left(3 x + 4\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(4 x + 3\right)^{6} e^{x} \sin^{4}{\left(x \right)}}{\left(- 6 x - 4\right)^{2} \left(3 x + 4\right)^{7}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = x + 6 \ln{\left(4 x + 3 \right)} + 4 \ln{\left(\sin{\left(x \right)} \right)}- 2 \ln{\left(- 6 x - 4 \right)} - 7 \ln{\left(3 x + 4 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = 1 + \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{24}{4 x + 3} - \frac{21}{3 x + 4} + \frac{12}{- 6 x - 4} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(1 + \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{24}{4 x + 3} - \frac{21}{3 x + 4} + \frac{12}{- 6 x - 4}\right)\left(\frac{\left(4 x + 3\right)^{6} e^{x} \sin^{4}{\left(x \right)}}{\left(- 6 x - 4\right)^{2} \left(3 x + 4\right)^{7}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(1 + \frac{4}{\tan{\left(x \right)}} + \frac{24}{4 x + 3}- \frac{21}{3 x + 4} + \frac{12}{- 6 x - 4}\right)\left(\frac{\left(4 x + 3\right)^{6} e^{x} \sin^{4}{\left(x \right)}}{\left(- 6 x - 4\right)^{2} \left(3 x + 4\right)^{7}} \right)