Using differentials, find the approximate values of the following:

Let us assume that

Also, let x = 25 so that x + Δx = 25.1

⇒ 25 + Δx = 25.1

∴ Δx = 0.1

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

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 = 0.1

⇒ Δf = (–0.004)(0.1)

∴ Δf = –0.0004

Now, we have f(25.1) = f(25) + Δf

⇒ f(25.1) = 0.2 – 0.0004

∴ f(15) = 0.1996

Thus,