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Use differentials or linearization to approximate the value shown (note each method will give the same result for your chosen x = a). (Round your answer to four decimal places.) (1.04) ln(1.04)
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\[\ln(1+x) \approx x\] 1.04 * ln(1.04) ~ 1.04 * 0.04 = 0.0416
Oh, I thought the function would be xlnx
and when I put that in it said it was wrong
1.04 in front of ln(1.04) is just a constant. You could write it as. f(x) = x * ln(x) ~ x * (x - 1) When x ~ 1
Hmm, okay so would I use differentials or linearization ?
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Hmm I'm not sure how they teach it in your school / textbook but I think that you actually use both. The linearization contains the differential. I'm not fully sure, wait for someone else to answer.
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