Using differentials, find the approximate value of `sqrt(49. 5)`

Explanation

Here, we will use the following rule,

`f(x+Deltax) = f(x) + f'(x)Deltax`

Here, `f(x+Deltax) = sqrt49.5, f(x) = sqrtx, x = 49, Deltax = 0.5`

`:. sqrt(x+Deltax) = sqrtx+d/dx(sqrtx)Deltax`

`:. sqrt(x+Deltax) = sqrtx+1/(2sqrtx)Deltax`

`=>sqrt(49.5) = sqrt49+1/(2sqrt49)(0.5)`

`=7+1/14(0.5) = 7+1/28~~ 7.0357`

`:. sqrt49.5 = 7.0357`