Why won't (x^2)^.5 differentiate right?!!!!! (need..help...)

GRRR!!! I am getting so sick of this little idiot. Let me tell you something. The Chain Rule must be a republican donor it is such a bastard. -_-

On that note.. someone tell me what I'm doing wrong...

y=(x^2)^.5 (aka square root of x-squared)

u=x^2
y=u^.5

dy/du=.5u^-.5
du/dx=2x

dy/du=.5(x^2)^-.5

dy/dx = dy/du * du/dx

dy/dx = .5(x^2)^-.5 * 2x

SO WHY ISN'T IT???

Maybe my calculator is graphing wrong and I was right all along.. but it looks more right than mine does... I wish it could tell me what it's equation is, but it can only show a graph.

...On looking over the post, I think I may have actually changed what I was doing to make it correct. Then again, maybe not. And my calculator is now downstairs, and if I went all that way and was still wrong, it might suffer a sudden acceleration into a wall. So I'm gonna just go ahead with asking for help first.

And should not be done outside of a computer program.

I don't even know where to start answering that one!

And then you differentiate that to 1, right?

Of course, I was never good with differential equations.

Funny, I actually was a math whiz until I hit differential equations (past regular old calculus)

so from

dy/dx = .5(x^2)^-.5 * 2x

You should get

x (x^2)^-.5

....whatever the hell that is. :)

You have a straight line with slope of 1.

not just x. Right?

Right, that's why you can't simplify it.

I wasn't thinking about negative numbers.

yeah, you have to technically split the domain into two (pos/neg) since square root is technically not a function

What's (x^2)^(1/2) ? Don't worry about thinking about square roots, and shit. Just expand out the brackets:

(x^2)^(1/2) = x^(2 x 1/2) = x^(2/2) = x

:shrug:

I never really understood differential equations, or how it was different from regular calculus, but I think for some reason you're not allowed to simplify the equation down, to begin with.

Part of why I got a D+ in that class- maybe I should have gone. :)

Not to mention that the clock says 4AM. But IMHO, that's it. Abs(x) isn't continuous, hence not derivable .

abs(x) is continuous (you can draw it without lifting your pencil).
it isn't differentiable at 0 since it isn't smooth

I didn't learn math in English.

You can draw it without lifting your pencil, but you can't draw its derivative without lifting.

the answer comes out as 1 and when you differentiate x you get 1.

undefined at x = 0

the u and y stuff is hard to remember. I remember it as:

"the derivative of the outside, OF the inside, times the derivative of the inside"
or like

"derivativeoutside(inside) x derivativeinside"

the derivative of the outside is 1/2 (inside)^-1/2, so the first part is
1/2 (x^2)^-1/2

then the derivative of the inside is 2x

so you have

<1/2 (x^2)^-1/2> * <2x>

simplifies to
(1/|x|)(x)

y is basically sqrt(x^2) or x, so dy/dx = 1

dy/dx as you have it is
= 1/2 * 1/sqrt(x^2) * 2x
= 2x / 2 sqrt(x^2)
= 2x / 2x
= 1

ex-calc TA - calculus can be used in ANYTHING (except masterbation - don't know how that comment got in this forum)

You might be the only poster whose first post was a diffeq problem. :toast:

Simplifying your equation, dy/dx = .5(x^2)^-.5 * 2x = .5(x^-1)*(2x) = 1

And that is the same result as you would get if you did it the easy way in the first place:

y=(x^2)^.5 = x

dy/dx = 1

Of course, you probably figured this all out by now, given that you have 20 more problems to figure out by tomorrow's class. :evilgrin:

on edit: I have been humbled by those who have trod before me...

which is the answer you calculated, so why do you think your answer is wrong? BTW when I use a TI Voyage 200 (functionally like a TI 89), I get dy/dx = sign(x), which agrees with the above (except sign(0) displays on the calculator as +-1).

some are right, some wrong, but they all sound certain! :)
The last answer is the correct one (I tutor calculus) -1 for x < 0, +1 x >0, and undefined x=0.

( x^2 ) ^1/2 = ( x^1/2 ) ^2 = x