Let us assume that

Also, let x = 36 so that x + Δx = 37

⇒ 36 + Δx = 37

∴ Δx = 1

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

We know

When x = 36, 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 = (0.08333)(1)

∴ Δf = 0.08333

Now, we have f(37) = f(36) + Δf

⇒ f(37) = 6 + 0.08333

∴ f(37) = 6.08333

Thus,