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

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.

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

Doesn't look right, should be 3*x |x|

dy/du = 3u^2, not 2u^2

-Uncool-

oops yeah i mean 3u^2 and the the answer is 3x|x| ???

hmmm i am still stuck on the "show"/prove bit of the problem??

 

?????????????????????

don't worry i think i got it! yay!

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?

defintion?