Using differential what is the approximate value of √(25.01)

Answer 1

We can use differential calculus to approximate the value of √(25.01) by finding the derivative of the function f(x) = √x at x = 25.

First, find the equation for the derivative of f(x):
f'(x) = 1 / (2√x)

Next, evaluate f'(25):
f'(25) = 1 / (2√25) = 1 / (2 * 5) = 1/10

Finally, use the linear approximation formula:
f(25.01) ≈ f(25) + f'(25)(25.01 - 25)
f(25.01) ≈ √25 + 1/10(25.01 - 25)
f(25.01) ≈ 5 + 0.1(0.01)
f(25.01) ≈ 5 + 0.001
f(25.01) ≈ 5.001

Therefore, the approximate value of √25.01 using differential calculus is 5.001.