calculus
posted by kendra .
use differentials to approximate the value of the squareroot of 4.3

calculus 
MathMate
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=(4.32^2)/(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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