DarkAngelBGE Posted March 23, 2004 Share Posted March 23, 2004 f(x) = sqrt( abs ( x ) )f'(x) = ? Link to comment Share on other sites More sharing options...
Jeru Posted March 23, 2004 Share Posted March 23, 2004 Does abs(x) mean |x|?If so, I don't know how to derive that function... It doesn't appear in my Calculus book either.In any case, here's half the solution:t = |x|(sqrt(t))' = t' / 2sqrt(t) = t' / 2sqrt(|x|) Link to comment Share on other sites More sharing options...
DarkAngelBGE Posted March 23, 2004 Author Share Posted March 23, 2004 abs(x) = |x| yes. Link to comment Share on other sites More sharing options...
MarkT Posted March 24, 2004 Share Posted March 24, 2004 dy/dx = sgn(x) / ( 2 * sqrt( |x| ) )f(x) = x^0.5 (x>0) = (-x)^0.5 (x<0)f'(x) = 0.5*x^-0.5 (x>0) = -0.5*|x|^-0.5 (x<0)= sgn(x) / 2 * x^-0.5= sgn(x) / ( 2 * x ^ 0.5 )= sgn(x) / ( 2 * sqrt( |x| ) ) Link to comment Share on other sites More sharing options...
Jeru Posted March 24, 2004 Share Posted March 24, 2004 dy/dx = sgn(x) / ( 2 * sqrt( |x| ) )f(x) = x^0.5 (x>0)Â Â Â = (-x)^0.5 (xf'(x) = 0.5*x^-0.5 (x>0)Â Â Â = -0.5*|x|^-0.5 (x= sgn(x) / 2 * x^-0.5= sgn(x) / ( 2 * x ^ 0.5 )= sgn(x) / ( 2 * sqrt( |x| ) ) sgn(x)... Now there's a nifty function.I was thinking about the derivative of |x| yesterday, and I knew it was 1 for x>0 and -1 for x Link to comment Share on other sites More sharing options...
DarkAngelBGE Posted March 24, 2004 Author Share Posted March 24, 2004 What does sgn( x ) stand for ?My solution is the following:f'(x) = 1 / ( 2 * sqrt( x ) ); x > 0f'(x) = -1 / (2 * sqrt( -x ) ); x < 0 Link to comment Share on other sites More sharing options...
MarkT Posted March 24, 2004 Share Posted March 24, 2004 sgn( x ) is defined as -1 for x<0, 1 for x>0 and 0 for x=0. It's also known as 'the sign function' or 'signum'.It has an approximately equivalent definition of x/|x| if you prefer it (gets a little strange for x=0 though) Link to comment Share on other sites More sharing options...
av_nefardec Posted March 25, 2004 Share Posted March 25, 2004 The derivative of the signum function is know as the floor function, is it not? Link to comment Share on other sites More sharing options...
DarkAngelBGE Posted March 25, 2004 Author Share Posted March 25, 2004 Ah alright cool. Link to comment Share on other sites More sharing options...
MarkT Posted March 25, 2004 Share Posted March 25, 2004 The derivative of the signum function is know as the floor function, is it not? I don't believe so, at least I've never seen the floor (or signum) functions defined so.I'd say that the derivative of sign/signum is the delta function. The floor function, as far as I know, is the greatest integer less than it's argument. It woud take a strange function to differentiate to that.By the by, here is a wonderful place to go for information on functions of all kinds. Very interesting stuff here. The two attached sites Mathworld and Scienceworld are effectively encyclopedias of mathematics and sciences (paritcularly physical sciences). They can be quite handy for doing homework. Link to comment Share on other sites More sharing options...
DarkAngelBGE Posted March 25, 2004 Author Share Posted March 25, 2004 Nice links, Mark, thanks. Link to comment Share on other sites More sharing options...
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