why is it? (1 Viewer)

[tacogym27101990]

why is it when you differentiate lnx, which is defined only for x>0 we get a function that is defined for all real x, except x=/=0??
probably a silly question but i was just wondering

 

[gurmies]

hmm, just a seemingly educated guess for me, but I assume that maybe you can think of it like this:

Suppose you have lnx, which is defined for x > 0, when u differentitate it, what do you get? You get the acceleration, which can be anything? As for the 0, I have no idea. I'm just throwing ideas out there

 

[Trebla]

why is it when you differentiate lnx, which is defined only for x>0 we get a function that is defined for all real x, except x=/=0??
probably a silly question but i was just wondering

The original statement is not quite true.
d(ln x)/dx = 1/x ONLY for x > 0.

Think about it. The graph of y = ln x is always increasing so its derivative is always positive and 1/x is only positive whenever x > 0, which of course is also the domain for y = ln x.

This derivative function of y = ln x is not the full hyperbola, it's only the positive branch. (i.e. y' = 1/x for x > 0), hence the derivative is actually only defined for x > 0 as well.

 

[tacogym27101990]

The original statement is not quite true.
d(ln x)/dx = 1/x ONLY for x > 0.

Think about it. The graph of y = ln x is always increasing so its derivative is always positive and 1/x is only positive whenever x > 0, which of course is also the domain for y = ln x.

This derivative function of y = ln x is not the full hyperbola, it's only the positive branch. (i.e. y' = 1/x for x > 0), hence the derivative is actually only defined for x > 0 as well.

yeah thats what i thought it would be

thanks