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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

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Topic review (newest first)

Ricky
2006-01-24 02:31:05

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

justlookingforthemoment
2006-01-23 17:48:31

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

irspow
2006-01-23 14:28:42

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.

ryos
2006-01-23 11:58:45

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.

irspow
2006-01-23 08:33:16

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.

suetonius
2006-01-23 06:30:12

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

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