Q. 9 J

# Using diffe

log1010.1, it being given that log10e = 0.4343

Let us assume that f(x) = log10x

Also, let x = 10 so that x + Δx = 10.1

10 + Δx = 10.1

Δx = 0.1

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 = 0.1

Δf = (0.04343)(0.1)

Δf = 0.004343

Now, we have f(10.1) = f(10) + Δf

f(10.1) = log1010 + 0.004343

f(10.1) = 1 + 0.004343 [ logaa = 1]

f(10.1) = 1.004343

Thus, log1010.1 ≈ 1.004343

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