Using differentials , find the approximate value of .

Let y = f(x) = √x

Let x = 0.49 and as we have to find for x = 0.62(that is f(0.62)) hence the small change dx will be 0.62 – 0.49 = 0.13

Using dy = f(x + dx) – f(x)

⇒ dy = f(0.49 + 0.13) – f(0.49)

⇒ dy = f(0.62) - √0.49 …since f(x) = √x

⇒ dy = f(0.62) – 0.7

⇒ f(0.62) = 0.7 + dy …(a)

Now we have to find dy

Differentiate y with respect to x

As x = 0.49 and dx = 0.13

Put this value of dy in (a)

⇒ f(0.62) = 0.792

Hence approximate value of √0.62 is 0.792