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