Question

use differential to approximate
$\sqrt{36.6}$
​

Answer 1

Step-by-step explanation:

meet your answer

Let y =√x

let x =36 & Δ x=0.6

Since y=√x

dy/dx =d(√x)/dx =1/2(√x)

Now:- Δy= dy/dx Δx

= 1/2(√x)(0.6)

=1/2√36(0.6

=0.6/2x6

=0.05

Also Δy= f(x+Δx)-f(x)

Putting value :

0.05=√36+0.6 -√36

√36.6=0.05+√36

√36.6=0.05+6

Δy=√x+Δx-√x

√36.6=6.05

Hence approximate value of √36.6 is 6.05