Can anybody help me find the antiderivative of 1/sqrt(x)?
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;
-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.
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.
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.
Well, I was going to move it, but then I wasn't sure whether suetonius would be able to find it again.
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..."