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LOL BUTTS LOL
1
Name:
Anonymous
2007-10-04 12:41
can anon derive f(x) = x^x ?
2
Name:
Anonymous
2007-10-04 12:45
f'(x)=x(x^(x-1))
3
Name:
Anonymous
2007-10-04 15:10
f(x)=x²
f'(x)=2x
4
Name:
Anonymous
2007-10-04 15:14
x^x = e^[xln(x)]
f(x) = e^[xln(x)]
f'(x) = (1+ln(x))x^x
5
Name:
Anonymous
2007-10-04 23:04
f(x) = x^x
ln f = x lnx
f'/f = (lnx - 1)/(lnx)^2
f' = (lnx - 1)(lnx)^2 * x^x
6
Name:
Anonymous
2007-10-04 23:15
>>2
Only works for constant exponents
>>3
x^x != x^2
>>4
Correct.
>>5
How did you get line 3 from line 2?
7
Name:
Anonymous
2007-10-05 2:09
>>6
He got there by implicit differentiation duh.
d/dx ln(f(x)) = 1/f df/dx
8
Name:
Anonymous
2007-10-05 12:23
>>6
Yay I winz!
>>7
No shit. But he did the product rule completely wrong.
I think he tried to use the quotient rule...
9
Name:
Anonymous
2007-10-05 16:22
Quotient rue is for pussies.
10
Name:
Anonymous
2007-10-06 12:18
http://calc101.com/webMathematica/derivatives.jsp
It solves derivatives for you so you don't have to ask anon and get the wrong answer.
Don't change these.
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