Thank you for explaining this!

To elaborate for anyone who’s curious: the square root function is 𝑓(𝑥) = 𝑥^½, so its derivative is 𝑓′(𝑥) = ½𝑥^-½ (power rule) = 1 / 2𝑥^½ = 1 / 2𝑓(𝑥).

If we know the value of 𝑓 for some nearby input 𝑎, we can approximate its value at 𝑥 by using 𝑓’s derivative (i.e. rate of change) at 𝑎 to guess how much the value of 𝑓(𝑎) has increased or decreased by the time it becomes 𝑓(𝑥). We can calculate that guess by multiplying 𝑓′(𝑎) by the difference between 𝑎 and 𝑥. It’s only a guess because it assumes that 𝑓 has a constant rate of change from 𝑎 to 𝑥, which the square root function doesn’t (it’s a curve, not a straight line), but if 𝑎 and 𝑥 are close together then the error isn’t too large.

e.g. 𝑓(67) ≈ 𝑓(64) + 𝑓′(64) × (67 - 64) = 8 + (1/16) × 3 = 8.1875