Use differentials to approximate ln 203 0992Solutionlet y f

Use differentials to approximate ln (2.03 - (0.99)^2).

let y= f(x) = ln x
f\'(x) = 1/x

x = 2.03
delta x = 0.99^2 = 0.9801

f\'(2.03) = 1/2.03

delta y = f\'(x)*delta x
= (1/2.03)*0.9801
=0.483

y= ln(x)
y(2.03)= ln (2.03) = 0.708

so,
ln (2.03 - (0.99)^2)
= y +-delta y
= 0.708 - 0.483
= 0.225

Answer: 0.225

$Use differentials to approximate ln (2.03 - (0.99)^2).Solutionlet y= f(x) = ln x f\'(x) = 1/x x = 2.03 delta x = 0.99^2 = 0.9801 f\'(2.03) = 1/2.03 delta y = f$

Use differentials to approximate ln 203 0992Solutionlet y f

Solution

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