Using differentials, find the approximate values of the following:

Let us assume that


Also, let x = 25 so that x + Δx = 25.1


25 + Δx = 25.1


Δx = 0.1


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








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 = 0.1


Δf = (0.004)(0.1)


Δf = 0.0004


Now, we have f(25.1) = f(25) + Δf




f(25.1) = 0.2 – 0.0004


f(15) = 0.1996


Thus,


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