Use differentials to approximate square root of 26

Answer 1

Let f(x) = √x. Then, f'(x) = 1/(2√x)

Let x = 25 and x+Δx = 26. So, Δx = 1

Δf = f(x+Δx) - f(x) = f(26) - f(25) = √26 - √25

df = f'(x)Δx = f'(25)(1) = 1/10

Δf ≈ df

So, √26 - √25 ≈ 1/10

√26 ≈ 5.1