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Find $ f'(a) $.

$ f(x) = x^{-2} $

$f^{\prime}(a)=-2 a^{-3}$

02:55

Daniel J.

01:08

Carson M.

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 7

Derivatives and Rates of Change

Limits

Derivatives

Campbell University

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So in this problem we are asked to find f prime at a when half of x is X to the -2. Well, by definition, if private A Is the limit as H approaches zero of F. Of a plus H minus F of a over H. All right. So that means that this is now the limit as h approaches zero F of a plus age and in this function F of X. So that is one over a plus H squared minus one over a squared all that is over eight. All right. So that means I have a limit as H goes to zero then of what? When I get a common denominator in the top. All right. And so I'll have a squared coming on top and then combine the fractions. That means I'll have a squared minus a squared plus two. A H plus H squared all over a squared times A plus H squared times H from the from the denominator previously. All right. So like this a squared minus at a square is but those two are gonna be gone, aren't they? So that means I have the limit as h goes to zero of minus to a H plus H squared over a squared times A plus h squared H. But I have an H in each term in the numerator so I can cancel one of those out. Can't I like that? So, that means I now have the limit As a joke goes to zero of minus to a plus H over a squared times A plus H squared. Okay, so taking now the limit as H goes to zero in the numerator, I'm just left with minus two way, aren't I? And the denominator. I have a squared times A squared, right? Because H goes to zero. So A plus H just goes to a square that let's just a square. So that means I'm left with -2 A over eight of the 4th. Well I can cancel 1, 1 of those AIDS out. and so that means I'm left with -2 over a cute is our derivative If prime it ain't.

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