Define the function and the first principles formula.

Let $f(x) = x^2 + \frac{3}{2}$.
The derivative from first principles is given by the formula:

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

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

$$f(x+h) = (x+h)^2 + \frac{3}{2}$$

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

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

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

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

Substitute the expression from Step 3 into the difference quotient:

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

Factor out $h$ from the numerator:

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

Cancel out $h$ (since $h \neq 0$ as $h$ approaches 0):

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

Apply the limit to the simplified difference quotient:

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

As $h$ approaches 0, the term $h$ becomes 0:

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

The derivative of $x^2 + \frac{3}{2}$ with respect to $x$ from first principles is $\boxed{\text{2x}}$.