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**suetonius****Member**- Registered: 2006-01-22
- Posts: 1

Can anybody help me find the antiderivative of 1/sqrt(x)?

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

The derivative of any fractional function:

f(x) = f(g)/f(h), f'(x) = [ f'(g)f(h) - f(g)f'(h) ] / (f(h))²

In your case;

f(g) = 1, f(h) = √x, f'(g) = 0, f'(h) = 1/(2√x)

So, using the formula for differentiating fractions above, this all becomes;

[ 0(√x) - 1(1/(2√x))] / (√x)²;

This equals;

-1/(2x√x) = (-1/2)x^(-3/2)

Note that this type of discussion belongs in the "HELP ME!" section. Please post your math questions there in the future.

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

He was asking for the antiderivative, AKA the integral.

1/√x can be rewritten as ½x^-½. This can be integrated simply by the power rule for integration: ∫½x^-½dx = x^½ or √x.

*Last edited by ryos (2006-01-22 13:01:30)*

El que pega primero pega dos veces.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

Haha, I guess that I was thrown off by this being posted in the wrong section. It seems fitting that a wrong answer was given for a question posted in the wrong place.

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**justlookingforthemoment****Moderator**- Registered: 2005-05-26
- Posts: 2,161

Well, I was going to move it, but then I wasn't sure whether suetonius would be able to find it again.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Move it, but don't delete this thread, and place a "Thread moved to here: " link for this thread.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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