Q. 9 A5.0( 1 Vote )
Using differentials, find the approximate values of the following:
Answer :
Let us assume that 
Also, let x = 25 so that x + Δx = 25.02
⇒ 25 + Δx = 25.02
∴ Δx = 0.02
On differentiating f(x) with respect to x, we get
We know
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 = 0.02
⇒ Δf = (0.1)(0.02)
∴ Δf = 0.002
Now, we have f(25.02) = f(25) + Δf
⇒ f(25.02) = 5 + 0.002
∴ f(25.02) = 5.002
Thus, 
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