Find the derivative of LaTeX: \displaystyle y = \frac{\left(5 x - 3\right)^{3} \sqrt{\left(4 x + 7\right)^{3}} e^{- x} \cos^{5}{\left(x \right)}}{\left(6 x + 1\right)^{7} \sin^{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(5 x - 3\right)^{3} \sqrt{\left(4 x + 7\right)^{3}} e^{- x} \cos^{5}{\left(x \right)}}{\left(6 x + 1\right)^{7} \sin^{4}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{3 \ln{\left(4 x + 7 \right)}}{2} + 3 \ln{\left(5 x - 3 \right)} + 5 \ln{\left(\cos{\left(x \right)} \right)}- x - 7 \ln{\left(6 x + 1 \right)} - 4 \ln{\left(\sin{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = - \frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 - \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{42}{6 x + 1} + \frac{15}{5 x - 3} + \frac{6}{4 x + 7} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(- \frac{5 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 - \frac{4 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{42}{6 x + 1} + \frac{15}{5 x - 3} + \frac{6}{4 x + 7}\right)\left(\frac{\left(5 x - 3\right)^{3} \sqrt{\left(4 x + 7\right)^{3}} e^{- x} \cos^{5}{\left(x \right)}}{\left(6 x + 1\right)^{7} \sin^{4}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 5 \tan{\left(x \right)} + \frac{15}{5 x - 3} + \frac{6}{4 x + 7}-1 - \frac{4}{\tan{\left(x \right)}} - \frac{42}{6 x + 1}\right)\left(\frac{\left(5 x - 3\right)^{3} \sqrt{\left(4 x + 7\right)^{3}} e^{- x} \cos^{5}{\left(x \right)}}{\left(6 x + 1\right)^{7} \sin^{4}{\left(x \right)}} \right)