Use differentials to obtain an approximation for (the square root of 50)

Question

we are supposed to solve this without a calculator...

Answer 1

Let f(x) = √x

 

Δf = f(x+Δx) - f(x) = √(x+Δx) - √x

 

df = differential of f = f'(x)Δx = 1/(2√x) Δx

 

If Δx is small, then Δf ≈ df

 

Let x = 49 and x+Δx = 50, then Δx = 1

 

So, Δf = √50 - √49  and df = 1/(2√49)(1) = 1/14

 

√50 - 7 ≈ 1/14

 

√50 ≈ 7 1/14