Using differentials, find the approximate values of the following:

Let us assume that


Also, let x = 0.09 so that x + Δx = 0.082


0.09 + Δx = 0.082


Δx = –0.008


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




We know





When x = 0.09, 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 = –0.008


Δf = (1.6667)(–0.008)


Δf = –0.013334


Now, we have f(0.082) = f(0.09) + Δf



f(0.082) = 0.3 – 0.013334


f(0.082) = 0.286666


Thus,


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