Use the First Principle to differentiate the following f(x) = 3x² − 4x + 2 and f(x) = sqrt(x)

$$f(x+h) = 3(x+h)^2 - 4(x+h) + 2$$ $$f(x+h) = 3(x^2 + 2xh + h^2) - 4x - 4h + 2$$ $$f(x+h) = 3x^2 + 6xh + 3h^2 - 4x - 4h + 2$$

$$f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 - 4x - 4h + 2) - (3x^2 - 4x + 2)$$ $$f(x+h) - f(x) = 3x^2 + 6xh + 3h^2 - 4x - 4h + 2 - 3x^2 + 4x - 2$$ $$f(x+h) - f(x) = 6xh + 3h^2 - 4h$$

$$\frac{f(x+h) - f(x)}{h} = \frac{6xh + 3h^2 - 4h}{h}$$ $$\frac{f(x+h) - f(x)}{h} = \frac{h(6x + 3h - 4)}{h}$$ $$\frac{f(x+h) - f(x)}{h} = 6x + 3h - 4$$

$$f'(x) = \lim_{h \to 0} (6x + 3h - 4)$$ $$f'(x) = 6x + 3(0) - 4$$ $$f'(x) = 6x - 4$$

The derivative of $f(x) = 3x^2 - 4x + 2$ is $\boxed{6x - 4}$.

2. Differentiate $f(x) = \sqrt{x}$

To evaluate this limit, multiply the numerator and denominator by the conjugate of the numerator.

$$f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \times \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}$$ $$f'(x) = \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}$$ $$f'(x) = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}$$

Cancel $h$ from the numerator and denominator:

$$f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$$

Substitute $h=0$:

$$f'(x) = \frac{1}{\sqrt{x+0} + \sqrt{x}}$$ $$f'(x) = \frac{1}{\sqrt{x} + \sqrt{x}}$$ $$f'(x) = \frac{1}{2\sqrt{x}}$$

The derivative of $f(x) = \sqrt{x}$ is $\boxed{\frac{1}{2\sqrt{x}}}$.