Using differentials, find the approximate value of √49.5

Let f(x) = √x, where x = 49
Let Δx = 0.5
Therefore, f(x + Δx) = √x + Δx = √49 + 0.5
Now by definition, approximately we can write,
f'(x) = (f(x + Δx) - f(x)) / Δx — (1)
Here,
f(x) = √x = √49 = 7
Δx = 0.5
f'(x) = 1 / (2√x) = 1 / (2√49) = 1/14
Putting these values in (1), we get
1/14 = (√49.5 - 7) / 0.5
=> √49.5 = 0.5/14 + 7 = 0.5 + 98/14 = 98.5/14 = 7.036
Therefore, the approximate value of √49.5 = 7.036