$\begin{align}  & y=\sin 3x \\ & y+\delta x=\sin 3(x+\delta x) \\ & \delta y=\sin (3x+3\delta x)-\sin 3x\left| note:\sin A-\sin B=2\cos \frac{A+B}{2}\sin \frac{A-B}{2} \right. \\ & \delta y=2\cos \frac{3x+3\delta x+3x}{2}\sin \frac{3x+3\delta x-3x}{2} \\ & \delta y=2\cos (3x+\tfrac{3\delta x}{2})\sin \tfrac{3\delta x}{2} \\ & \text{Divide through by }\delta x \\ & \frac{\delta y}{\delta x}=\frac{2\cos (3x+\tfrac{3\delta x}{2})}{1}\cdot \frac{\sin \tfrac{3\delta x}{2}}{\delta x} \\ & \frac{\delta y}{\delta x}=\frac{2\cos (3x+\tfrac{3\delta x}{2})}{1}\cdot \frac{\sin \tfrac{3\delta x}{2}}{\tfrac{3\delta x}{2}}\times 3 \\ & \text{Note: This step we have it in the form of }\frac{\sin \theta }{\theta } \\ & \text{proceeding to the limit as }\delta x\to 0,\text{ we get} \\ & \frac{dy}{dx}=\underset{\delta x\to \infty }{\mathop{\lim }}\,\frac{\delta y}{\delta x}=3\cdot \underset{\delta x\to \infty }{\mathop{\lim }}\,\frac{2\cos (3x+\tfrac{3\delta x}{2})}{1}\cdot \underset{\delta x\to \infty }{\mathop{\lim }}\,\frac{\sin \tfrac{3\delta x}{2}}{\tfrac{3\delta x}{2}} \\ & \left| note: \right.\underset{\delta x\to \infty }{\mathop{\lim }}\,\frac{\sin \tfrac{3\delta x}{2}}{\tfrac{3\delta x}{2}}=1 \\ & \frac{dy}{dx}=3\times \cos 3x\times 1=3\cos 3x \\ & \text{hence,}\frac{d}{dx}(\sin 3x)=3\cos 3x \\\end{align}$