You are not logged in.
#2 2006-01-23 08:33:16
- irspow
- Power Member
Offline
Re: anitderivative 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;
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.
#3 2006-01-23 11:58:45
- ryos
- Power Member
Offline
Re: anitderivative of 1/sqrt(x)
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 (2006-01-23 12:01:30)
El que pega primero pega dos veces.
#5 2006-01-23 17:48:31
- justlookingforthemoment
- Moderator
-
Offline
Re: anitderivative of 1/sqrt(x)
Well, I was going to move it, but then I wasn't sure whether suetonius would be able to find it again.
#6 2006-01-24 02:31:05
- Ricky
- Moderator
Offline
Re: anitderivative of 1/sqrt(x)
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..."