Why is $d(x^2)= 2xdx$?

When treating the differential operator like a fraction, you might see this: $$\frac d {\color{blue}{dx}}(x^2)=2x\implies d(x^2)=2x\,\color{blue}{dx}.$$

We have $\Delta(x^2) = 2x \Delta x + (\Delta x)^2$. Note that $x^2 = y$ so, we have $\Delta(x^2) = \Delta y$. Thus, we have $\Delta y = 2x\Delta x +(\Delta x)^2$. Divide both sides by $\Delta x$ and we get $\frac{\Delta y}{\Delta x} = 2x + \Delta x$. Now, we take the limit as $\Delta x$ approaches $0$, giving us $dy/dx = 2x$. Note that $dy/dx$ is the limit as $\Delta x$ goes to $0$ of $\Delta y/\Delta x$ by definition.