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


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