$\begin{align}  & y={{\sin }^{2}}5x \\ & \text{Let }u=5x \\ & y={{\sin }^{2}}u={{(\sin u)}^{2}} \\ & \frac{dy}{du}=2\sin u\times (\cos u)\text{   }\left| chain's\text{ }rule \right. \\ & \frac{dy}{du}=2\sin u\cos u \\ & \frac{du}{dx}=\frac{d}{dx}(5x)=5 \\ & \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}=2\sin u\cos u\times 5 \\ & \frac{dy}{dx}=10\sin u\cos u=10\sin 5x\cos 5x \\\end{align}$