Originally Posted by

**Zabulius** I'm going to write this problem word for word because I'm not sure how else to describe it.

Q1. For y=f(x), differentials are used to approximate the change in y for a given change in x. Suppose you have misplaced your calculator and need a decent approximation of the square root of 27. The yknow 25^1/2 = 5 so you assume that 27^1/2 must be slightly larger than 5.

A. Use differentials to approximate the square root of 27.

Step 1. f(x) = x^1/2, x=25, dx=2.

Calculate dy = f '(x) dx and evaluate it.

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In addition to doing it for 27, I have to do it for 21, 39, and 110. The main problem is I don't know which values to put where. I think the x is always 25 and only dx changes? So for 27, dy = .5(25^.50)*2 which is .2 I believe. So my approximation would be 5.2? Any help or insight is appreciated. Thanks