Calculus with Applications (10th Edition)

$$\frac{{dy}}{{dx}} = \frac{{4\cos \left( {\ln 3{x^4}} \right)}}{x}$$
\eqalign{ & y = \sin \left( {\ln 3{x^4}} \right) \cr & {\text{differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = {D_x}\left[ {\sin \left( {\ln 3{x^4}} \right)} \right] \cr & {\text{use the chain rule }}{D_x}\left( {\sin u} \right) = \cos u \cdot {D_x}\left( u \right).{\text{ consider }}u = \ln 3{x^4} \cr & \frac{{dy}}{{dx}} = \cos \left( {\ln 3{x^4}} \right) \cdot {D_x}\left( {\ln 3{x^4}} \right) \cr & {\text{use the chain rule }}{D_x}\left( {\ln u} \right) = \frac{1}{u} \cdot {D_x}\left( u \right) \cr & \frac{{dy}}{{dx}} = \cos \left( {\ln 3{x^4}} \right)\left( {\frac{1}{{3{x^4}}}} \right) \cdot {D_x}\left( {3{x^4}} \right) \cr & {\text{solve the derivative and simplify}} \cr & \frac{{dy}}{{dx}} = \cos \left( {\ln 3{x^4}} \right)\left( {\frac{1}{{3{x^4}}}} \right)\left( {12{x^3}} \right) \cr & \frac{{dy}}{{dx}} = \cos \left( {\ln 3{x^4}} \right)\left( {\frac{4}{x}} \right) \cr & \frac{{dy}}{{dx}} = \frac{{4\cos \left( {\ln 3{x^4}} \right)}}{x} \cr}