Write down the definition of the derivative from first principles.

The derivative of a function $f(x)$ from first principles is given by:

$$\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

Given the function $f(x) = 2x^2$.
First, we need to find $f(x+h)$ by replacing $x$ with $(x+h)$ in the function:

$$f(x+h) = 2(x+h)^2$$

Expand the term $(x+h)^2$:

$$(x+h)^2 = x^2 + 2xh + h^2$$

Substitute this back into the expression for $f(x+h)$:

$$f(x+h) = 2(x^2 + 2xh + h^2) = 2x^2 + 4xh + 2h^2$$

Subtract $f(x)$ from $f(x+h)$:

$$f(x+h) - f(x) = (2x^2 + 4xh + 2h^2) - (2x^2)$$ $$f(x+h) - f(x) = 4xh + 2h^2$$

Now, divide the result from Step 2 by $h$:

$$\frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^2}{h}$$

Factor out $h$ from the numerator:

$$\frac{h(4x + 2h)}{h}$$

Cancel out $h$ from the numerator and the denominator (since $h \neq 0$ as $h$ approaches 0):

$$4x + 2h$$

Finally, apply the limit as $h$ approaches 0 to the simplified expression:

$$\frac{dy}{dx} = \lim_{h \to 0} (4x + 2h)$$

Substitute $h=0$ into the expression:

$$\frac{dy}{dx} = 4x + 2(0)$$ $$\frac{dy}{dx} = 4x$$

The differential coefficient is $\mathbf{4x}$.