Using differentials, find the approximate values of the following:

(15)1/4

Let us assume that


Also, let x = 16 so that x + Δx = 15


16 + Δx = 15


Δx = –1


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



We know





When x = 16, 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 = (0.03125)(1)


Δf = 0.03125


Now, we have f(15) = f(16) + Δf




f(15) = 2 – 0.03125


f(15) = 1.96875


Thus, (15)1/4 ≈ 1.96875


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