Posted by **kendra** on Wednesday, April 18, 2012 at 11:33am.

use differentials to approximate the value of the squareroot of 4.3

- calculus -
**MathMate**, Wednesday, April 18, 2012 at 5:21pm
Let

f(x)=sqrt(x)=x^(1/2)

f'(x)=1/(2sqrt(x))

and

f(x0+Δx)=f(x0)+f'(x0)*Δx (approx.)

or

Δx=(f(x0+Δx)-f(x0))/f'(x0) approx.

Knowing that 2^2=4, and 2.1^2=4.41

Try x0=2

Δx=(4.3-2^2)/(1/(2*2)

=0.3/(4)

=0.075

or

Approximately, x+Δx=2+.075=2.075

(check: 2.075^2=4.305625)

Try x0=2.1, x0^2=4.41...

to get a still better approximation.

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