Question

What is the derivative of x^(x^(1/2))?

Answer 1

I put it in Wolfram Alpha and this is what they gave me, but I have no idea how they got it: 1/2x^(x^(1/2)-1/2)(log(x) + 2)

Answer 2

is it equal to something. I want to use ln. If you times it by the natural log you can bring the power out in front and use the product rule to differentiate

Answer 3

thats probably what they did

Answer 4

so. (x^1/2)( ln x)

Answer 5

The actual problem is: Find the limit of that equation as x goes to 0+.

Answer 6

What calc is this

Answer 7

Calculus 2. This is an indeterminate form problem

Answer 8

You end up getting 0^(0) which requires L'hospital's rule

Answer 9

Which requires taking the derivative of that :(

Answer 10

The limit appears to go to 1.

Answer 11

How did you get that?

Answer 12

I can find the derivative but honestly I can't remember L'hospital's rule. I vaguely remember taking notes on this years ago. Hopefully somebody else can help. I don't want to steer you in the wrong direction

Answer 13

I'm not sure L'Hopital's would help here since it's not a ratio. It's not like a 0/0 or ∞/∞.

Answer 14

Cliffsedge, L'hopital's rule is done when you get: 00, 0/0, 1∞, ∞ − ∞, ∞/∞, 0 × ∞, and ∞0. (from Wikipedia)

Answer 15

So in this case, you get 0^0 which is one of the known ones :/.

Answer 16

Understood, but what is the process of L'Hopital's Rule?

Answer 17

L'Hopital's rule says that if you get in a situation like this, you take the derivative of it. F(x)/g(x) becomes f'(x) / g'(x)

Answer 18

For mine, since we don't have a fraction, we just take the derivative, and see if plugging 0 again would give us a situation like the ones I listed. If I do, I use it again, if I don't, then whatever I get is my final answer.

Answer 19

If I keep getting loops, and keep getting stuck, and have no other choices to do anything, then it's an undefined problem.

Answer 20

I don't think you can just take the derivative right off like that. You can exponentiate first, then move the limit inside the exponent.

Answer 21

hmm, the professor does very simple examples in class -.-

Answer 22

\[x^{\sqrt{x}} \rightarrow e^{\ln(x^{\sqrt{x}})} \rightarrow e^{\sqrt{x}\ln(x)} \]

Answer 23

hmm I think that's right, but it requires another L'hopital's rule, because I just plugged in 0, and it doesn't work

Answer 24

Yes, now you can find
\[\lim_{x \rightarrow 0+}\sqrt{x}\ln(x) = \lim_{x \rightarrow 0+}\frac{\ln(x)}{x^{-1/2}}\]

Answer 25

Now regular L'Hopital's rule will break that one down and you should see that it goes to zero.

Answer 26

What happened to the e?

Answer 27

It's still there. I just moved the limit evaluation to the exponent to show that it is e^0 =1.

Answer 28

hmm now if I plug in 0, it still does the same thing.

Answer 29

If you still get 0/0, use L'Hopital's again until the denominator goes to 1.

Answer 30

the bottom becomes -1/2x^(1/2)

Answer 31

Check your power rule again.

Answer 32

Hmm. x^(-1/2) = -1/2x^(1/2) * (1), no?

Answer 33

subtract one..ahh I was adding lol

Answer 34

I have been doing integrals all day long, so I just confused myself even more.
-1/2x^(-3/2)

Answer 35

\[\frac{d}{dx}x^{-1/2} = -\frac{1}{2}x^{-3/2}\]

Answer 36

yeah

Answer 37

I feel 'ya. That happens to me all the time. :-p

Answer 38

xD

Answer 39

Just keep simplifying and differentiating top and bottom until you get -3x^(1/2) over 1.

Answer 40

we have never used l'hopital's rule this many times o.o. But okay

Answer 41

Yeah, usually it does its thing in one or two applications, but sometimes you have to keep putting those functions through the wringer until they come out nice and smooth the way you want them.

Answer 42

okay, I will keep doing it. Thank you so much :)

Answer 43

You're welcome. This was a nasty one. Similar kind of annoyance as having to do integration-by-parts over and over again.

Answer 44

I think that's why I didn't understand it, integration by parts is actually next chapter's material..and the professor gave us some questions that require it, although we don't know it yet lol.

Answer 45

Some professors think you need to learn to fly before you learn to walk . . .