Using differentials, find the approximate values of the following:
(1.999)5
Let us assume that f(x) = x5
Also, let x = 2 so that x + Δx = 1.999
⇒ 2 + Δx = 1.999
∴ Δx = –0.001
On differentiating f(x) with respect to x, we get
We know
When x = 2, 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 = –0.001
⇒ Δf = (80)(–0.001)
∴ Δf = –0.08
Now, we have f(1.999) = f(2) + Δf
⇒ f(1.999) = 25 – 0.08
⇒ f(1.999) = 32 – 0.08
∴ f(1.999) = 31.92
Thus, (1.999)5 ≈ 31.92