Using differentials, find the approximate values of the following:

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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