Using differentials, find the approximate values of the following:

(255)^1/4

Let us assume that

Also, let x = 256 so that x + Δx = 255

⇒ 256 + Δx = 255

∴ Δx = –1

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

We know

When x = 256, 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 = (0.00390625)(–1)

∴ Δf = –0.00390625

Now, we have f(255) = f(256) + Δf

⇒ f(255) = 4 – 0.00390625

∴ f(255) = 3.99609375

Thus, (255)^1/4 ≈ 3.99609375