Let us assume that

Also, let x = 25 so that x + Δx = 26

⇒ 25 + Δx = 26

∴ Δx = 1

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

We know

When x = 25, we have

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

Here, and Δx = 1

⇒ Δf = (0.1)(1)

∴ Δf = 0.1

Now, we have f(26) = f(25) + Δf

⇒ f(26) = 5 + 0.1

∴ f(26) = 5.1

Thus,