Differentiate f(x) = 2x2 − 16 from the first principle.

Substitute $(x+h)$ into the function $f(x)$.

$$f(x+h) = 2(x+h)^2 - 16$$

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

$$f(x+h) = 2(x^2 + 2xh + h^2) - 16$$ $$f(x+h) = 2x^2 + 4xh + 2h^2 - 16$$

Subtract the original function $f(x)$ from $f(x+h)$.

$$f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 - 16) - (2x^2 - 16)$$ $$f(x+h) - f(x) = 2x^2 + 4xh + 2h^2 - 16 - 2x^2 + 16$$ $$f(x+h) - f(x) = 4xh + 2h^2$$

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:

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

Cancel $h$ (since $h \neq 0$ in the limit process):

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

Apply the limit to the expression from Step 3.

$$f'(x) = \lim_{h \to 0} (4x + 2h)$$

Substitute $h=0$ into the expression:

$$f'(x) = 4x + 2(0)$$ $$f'(x) = 4x$$

The derivative of $f(x) = 2x^2 - 16$ from the first principle is $\boxed{4x}$.