Using differentials, find the approximate values of the following:​​​​​​​ √25.02

Question

Using differentials, find the approximate values of the following:
\(\sqrt{25.02}\)

Answer 1

Let us assume that f(x) = \(\sqrt{x}\)

Also, let x = 25 so that x + Δx = 25.02

⇒ 25 + Δx = 25.02

∴ Δx = 0.02

On differentiating f(x) with respect to x, we get

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as