Use differentials to approximate the quantity.

Question

Use differentials to approximate the quantity.

30.0001246

Answer 1

Hi Elizabeth G.,

I'll assume you know what differentials are, and so on. So what is the function you are being asked to differentiate? It's simply f(x) = x^(1/3) , around the location f(x) = 0.0001246 . Now, solve the inverse function: g(x) = x^3, right? So what cubed is ∼ 0.000125? Right, so x ∼ 0.05 .... so write your expression for f(x-Δx) = f(x) - Δx * f ' (x) and plug in. OK? I'll leave it for you to find f ' (x) .

-- Cheers, -- Mr. d.

Answer 2

Let y = f(x) = x^1/3, x = 0.000125, x + Δx = 0.0001246. So, Δx = dx = -0.0000004

(0.0001246)^1/3 ≈ f(x) + f'(x)dx = (0.000125)^1/3 + [1 / (3(0.000125))^2/3](-0.0000004)

= 0.05 + [1/0.0075](-0.0000004) = 0.049946666

Note: 0.000125 was chosen as x because we can find the exact value of f(0.000125) and 0.000125 is "close" to 0.0001246.