Question

using differential find approximate valu of (0.036)^1/2

Answer 1

Answer:

(0.036)^(1/2)  ≈  0.19   (approx)

Step-by-step explanation:

Put f(x) = √x = x^(1/2)

Then f' (x) = (1/2) x^(-1/2) = 1 / 2√x

Now

f' (x) ≈ ( f(x+h) - f(x) ) / h   =>  f(x+h) ≈ f(x) + h f' (x)

The idea is to write 0.036 as x+h where we can evaluate f(x) and f' (x).  For this, we need a nearby x for which we know the square root.

Since 0.18² = 0.0324, this looks like a good candidate.

Take x = 0.0324 and h = 0.036 - 0.0324 = 0.0036.

Then

(0.036)^(1/2)

= f(0.036)

= f(x+h)

≈ f(x) + h f' (x)   (approximately)

= f(0.0324) + 0.0036 × f' (0.0324)

= √0.0324 + 0.0036 × 1 / 2√0.0324

= 0.18 + 0.0036 / ( 2 × 0.18 )

= 0.18 + 0.0036 / 0.36

= 0.18 + 0.01

= 0.19