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

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 x - 5\right)^{4} \sqrt{\left(2 x + 3\right)^{7}} \sin^{5}{\left(x \right)}}{\left(5 - 4 x\right)^{7} \left(x - 3\right)^{4}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{7 \ln{\left(2 x + 3 \right)}}{2} + 4 \ln{\left(5 x - 5 \right)} + 5 \ln{\left(\sin{\left(x \right)} \right)}- 7 \ln{\left(5 - 4 x \right)} - 4 \ln{\left(x - 3 \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{20}{5 x - 5} + \frac{7}{2 x + 3} - \frac{4}{x - 3} + \frac{28}{5 - 4 x} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{20}{5 x - 5} + \frac{7}{2 x + 3} - \frac{4}{x - 3} + \frac{28}{5 - 4 x}\right)\left(\frac{\left(5 x - 5\right)^{4} \sqrt{\left(2 x + 3\right)^{7}} \sin^{5}{\left(x \right)}}{\left(5 - 4 x\right)^{7} \left(x - 3\right)^{4}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{5}{\tan{\left(x \right)}} + \frac{20}{5 x - 5} + \frac{7}{2 x + 3}- \frac{4}{x - 3} + \frac{28}{5 - 4 x}\right)\left(\frac{\left(5 x - 5\right)^{4} \sqrt{\left(2 x + 3\right)^{7}} \sin^{5}{\left(x \right)}}{\left(5 - 4 x\right)^{7} \left(x - 3\right)^{4}} \right)