Using differentials, find the approximate values of the following:
Let us assume that
Also, let x = 25 so that x + Δx = 25.1
⇒ 25 + Δx = 25.1
∴ Δx = 0.1
On differentiating f(x) with respect to x, we get
When x = 25, 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 = 0.1
⇒ Δf = (–0.004)(0.1)
∴ Δf = –0.0004
Now, we have f(25.1) = f(25) + Δf
⇒ f(25.1) = 0.2 – 0.0004
∴ f(15) = 0.1996
Thus,