TeX джерело:

y' = (\sqrt {{x^2} + 2x} )' = \frac{1}{{2\sqrt {{x^2} + 2x} }} \cdot ({x^2} + 2x)' = \frac{{2x + 2}}{{2\sqrt {{x^2} + 2x} }} = \frac{{x + 1}}{{\sqrt {{x^2} + 2x} }}