Find the derivative of LaTeX: \displaystyle y = \frac{\sqrt{\left(4 x + 3\right)^{5}} \cos^{3}{\left(x \right)}}{64 x^{6} \left(x - 7\right)^{3} \sin^{5}{\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{\sqrt{\left(4 x + 3\right)^{5}} \cos^{3}{\left(x \right)}}{64 x^{6} \left(x - 7\right)^{3} \sin^{5}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = \frac{5 \ln{\left(4 x + 3 \right)}}{2} + 3 \ln{\left(\cos{\left(x \right)} \right)}- 6 \ln{\left(x \right)} - 3 \ln{\left(x - 7 \right)} - 5 \ln{\left(\sin{\left(x \right)} \right)} - 6 \ln{\left(2 \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)}} - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{10}{4 x + 3} - \frac{3}{x - 7} - \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)}} - \frac{5 \cos{\left(x \right)}}{\sin{\left(x \right)}} + \frac{10}{4 x + 3} - \frac{3}{x - 7} - \frac{6}{x}\right)\left(\frac{\sqrt{\left(4 x + 3\right)^{5}} \cos^{3}{\left(x \right)}}{64 x^{6} \left(x - 7\right)^{3} \sin^{5}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(- 3 \tan{\left(x \right)} + \frac{10}{4 x + 3}- \frac{5}{\tan{\left(x \right)}} - \frac{3}{x - 7} - \frac{6}{x}\right)\left(\frac{\sqrt{\left(4 x + 3\right)^{5}} \cos^{3}{\left(x \right)}}{64 x^{6} \left(x - 7\right)^{3} \sin^{5}{\left(x \right)}} \right)