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Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative.

$ f(x) = x^{3/2} $

$\frac{3}{2} x^{1 / 2}$

05:42

Daniel J.

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 8

The Derivative as a Function

Limits

Derivatives

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

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So in this problem were given this function F of X is X to the To the 3/2. And were asked to do several things. Were asked to find the domain of this, the derivative of using the function and the domain of the derivative. So, I want to start with the domain here, first of all. So the domain is 02 infinity. As I cannot do a negative number To the 1/2 power. Okay. So there's my domain for my function. Now, definition of derivative definition derivative says F prime of X is equal to the limit as H goes to zero of F of X plus H minus F of X over H. All right. So for us then that means derogative fx is the limit as H goes to zero of X plus H two, the three house minus X to the three house all over H. And we can see real quick is that if we multiply this by X plus H two the three halves plus X to the three house over X plus H three halves plus X to the three halves. Right? Because what are we doing? We're doing a let's see, we're doing a minus B times A plus B equals a squared minus b squared. All right. So that means this is the limit. His page goes to zero of x plus age to the three halves squared. Well, the one half power squared cancels out. And so I'm left with X plus H cubed minus X cubed, aren't I? Because extra three halves squared would be Next to the 3rd. Okay. Age times X plus H. 23 halves plus X. To the three halves. All right now multiply. This out Limit is H goes to zero. Uh X cubed plus three X square at H plus three X. H squared plus H cubed minus X cubed over. H times X plus H. Two. The three halves Plus X. 2 3/2. All right. And what do we notice? First of all we noticed that X cubed minus X cubed. So that's gone next. We noticed that I got an H. Here in the denominator and H. And that term and H there and I can take one of those ages cancel the H. Is out. Okay And so I'm left with what I'm left with the limit as H goes to zero of three x squared Plus three XH plus H squared over X plus H. Two the three halves plus X to the three halves. Okay now perform the limit Well when h goes to zero that term goes to zero that term goes to zero and that H goes to zero doesn't it? Some left with three X squared over execute was X cubed. Extra three halves plus extra three halves. That's two X. to the three. Well I didn't want to write very well let's try that again. X. to the three house. So that's three over to well X squared over extra three halves. That's X. To the two minus three halves in the export up there. So that leaves me X to the one half, doesn't it? Okay. And the domain then is still zero to infinity. Just like we said above. Right, We can't do a negative number to the one half, unless we can't do the square root of a negative number. So here is my derivative and its domain.

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