Question

Using differentials, find the approximate value of \( \sqrt {25.2} \)

Answer 1

Hope this helppp.............

Answer 2

Approximate value of √25.2 = 5.02

• For finding the approximate value of √25.2

Let y = f(x) = √x

Here, x = 25 and ∆x = 0.2 - (1)

• As y = √x

Therefore dy/dx = 1/2√x - (2)

• Now, ∆y = dy/dx (∆x)

Putting the value of dy/dx from (2) in the above equation, we get

∆y = 1/2√x (∆x) - (3)

• Now,

∆y = f(x + ∆x) - f(x) - (4)

Equating (3) and (4), we get

f(x + ∆x) - f(x) = 1/2√x (∆x)

f(x + ∆x) = 1/2√x (∆x) + f(x)

• Now putting the values of x , ∆x from (1) in the above equation we get,

f( 25 + 0.2) = 1/2√25 (0.2) + f(25)

f(25.2) = 1/10 (0.2) + √25

√25.2 = 0.2/10 + 5

√25.2 = 0.02 + 5

Therefore √25.2 = 5.02