Using differential, find the approximate value of ✓(25.01)

Answer 1

Let f(x) = ✓(x)

Using the definition of a derivative, we have:

f'(x) = lim(h->0) [f(x + h) - f(x)] / h

Let's approximate the value of f'(25.01) by using a small h value:

f'(25.01) ≈ [f(25.01 + 0.01) - f(25.01)] / 0.01
f'(25.01) ≈ [✓(25.02) - ✓(25.01)] / 0.01

By calculating these values (considering ✓(25.02) ≈ 5.002), we get:

f'(25.01) ≈ (5.002 - ✓(25.01)) / 0.01
f'(25.01) ≈ (5.002 - 5) / 0.01
f'(25.01) ≈ 0.002 / 0.01
f'(25.01) ≈ 0.2

Therefore, the approximate value of f'(25.01) is 0.2.