Find the derivative of LaTeX: \displaystyle y = \frac{\left(5 - 6 x\right)^{4} \sin^{7}{\left(x \right)}}{\left(2 x + 4\right)^{5} \sqrt{\left(5 x + 7\right)^{3}}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(5 - 6 x\right)^{4} \sin^{7}{\left(x \right)}}{\left(2 x + 4\right)^{5} \sqrt{\left(5 x + 7\right)^{3}}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 4 \ln{\left(5 - 6 x \right)} + 7 \ln{\left(\sin{\left(x \right)} \right)}- 5 \ln{\left(2 x + 4 \right)} - \frac{3 \ln{\left(5 x + 7 \right)}}{2} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{2 \left(5 x + 7\right)} - \frac{10}{2 x + 4} - \frac{24}{5 - 6 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{2 \left(5 x + 7\right)} - \frac{10}{2 x + 4} - \frac{24}{5 - 6 x}\right)\left(\frac{\left(5 - 6 x\right)^{4} \sin^{7}{\left(x \right)}}{\left(2 x + 4\right)^{5} \sqrt{\left(5 x + 7\right)^{3}}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{7}{\tan{\left(x \right)}} - \frac{24}{5 - 6 x}- \frac{15}{2 \left(5 x + 7\right)} - \frac{10}{2 x + 4}\right)\left(\frac{\left(5 - 6 x\right)^{4} \sin^{7}{\left(x \right)}}{\left(2 x + 4\right)^{5} \sqrt{\left(5 x + 7\right)^{3}}} \right)