Sarahisme Posted April 20, 2005 Share Posted April 20, 2005 hey again Show that f(x) = |x|^3 is differentiable at every real x, and find it’s derivative. i am not sure exactly what the question means by "show".... i got f'(x) = 3x.|x| (not even sure if that right but anyways) and i suppose that f'(x) is defined for all R numbers of x.... however i dont think thats really "showing" anything anyones thoughts on the matter would be welcomed Cheers Sarah Link to comment Share on other sites More sharing options...
Dave Posted April 20, 2005 Share Posted April 20, 2005 When they say "show", it's usually equivalent and right to substitute the word "prove" How do we know if something is differentiable? Well, it should obviously be continuous, and moreover, the limit: [math]\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}[/math] should exist. Maybe the fact that [math]|x| = \sqrt{x^2}[/math] would help. Link to comment Share on other sites More sharing options...
Sarahisme Posted April 22, 2005 Author Share Posted April 22, 2005 ok ignoring the "show for all x" bit, and just concentrating on finding the dervative of f(x) = |x|^3 this is how i calculatd it.....please tell me if i correct or incorrect .......... y = u^3 u = |x| dy/du = 2u^2 du/dx = x/|x| = sgn(x) dy/dx = dy/du * du/dx = (2u^2)*(x/|x|) = (2|x|^2)*(x/|x|) = 2x|x| ??? Thanks Sarah Link to comment Share on other sites More sharing options...
uncool Posted April 22, 2005 Share Posted April 22, 2005 Doesn't look right, should be 3*x |x| dy/du = 3u^2, not 2u^2 -Uncool- Link to comment Share on other sites More sharing options...
Sarahisme Posted April 24, 2005 Author Share Posted April 24, 2005 oops yeah i mean 3u^2 and the the answer is 3x|x| ??? Link to comment Share on other sites More sharing options...
Sarahisme Posted April 30, 2005 Author Share Posted April 30, 2005 hmmm i am still stuck on the "show"/prove bit of the problem?? ????????????????????? Link to comment Share on other sites More sharing options...
Sarahisme Posted May 1, 2005 Author Share Posted May 1, 2005 don't worry i think i got it! yay! Link to comment Share on other sites More sharing options...
Johnny5 Posted May 3, 2005 Share Posted May 3, 2005 ok ignoring the "show for all x" bit' date=' and just concentrating on finding the dervative of f(x) = |x|^3 this is how i calculatd it.....please tell me if i correct or incorrect .......... y = u^3 u = |x| dy/du = 3u^2 du/dx = x/|x| = sgn(x) dy/dx = dy/du * du/dx = (3u^2)*(x/|x|) = (3|x|^2)*(x/|x|) = 3x|x| ??? Thanks Sarah [/quote'] Is that signum function up there? If it is, what is its definition again? Link to comment Share on other sites More sharing options...
Sarahisme Posted May 5, 2005 Author Share Posted May 5, 2005 defintion? Link to comment Share on other sites More sharing options...
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