Q. 9 I

Using differentials, find the approximate values of the following:

loge10.02, it being given that loge10 = 2.3026

Answer :

loge10.02, it being given that loge10 = 2.3026


Let us assume that f(x) = logex


Also, let x = 10 so that x + Δx = 10.02


10 + Δx = 10.02


Δx = 0.02


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



We know



When x = 10, 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(10.02) = f(10) + Δf


f(10.02) = loge10 + 0.002


f(10.02) = 2.3026 + 0.002


f(10.02) = 2.3046


Thus, loge10.02 ≈ 2.3046


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