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

[Goldom]

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.

[Ediacara]

1. Calculus = Mathematical Masturbation

And should not be done outside of a computer program.

[billyskank]

24. !!!

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

[Guaranteed]

2. The square root of x squared...isn't that just x?

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)

[s-cubed]

9. No, the square root of x^2 is the absolute value of x

[Guaranteed]

11. Yeah, you're right,

so from

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

You should get

x (x^2)^-.5

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

[benburch]

3. y=x

You have a straight line with slope of 1.

[Goldom]

6. But square root of x^2 is the absolute value of x

not just x. Right?

[Guaranteed]

8. Yes, that's right.

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

[benburch]

14. Yes, true.

I wasn't thinking about negative numbers.

[LiberalBadger]

16. split the domain

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

[billyskank]

25. Look at it this way

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

[Kellanved]

4. 1?

:shrug:

[Guaranteed]

7. Yeah, I think this is different, for some reason...

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. :)

[Kellanved]

12. has been some time

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

[LiberalBadger]

18. huh?

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

[Kellanved]

22. language issue

I didn't learn math in English.

[billyskank]

26. Not *piecewise* continuous

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

[Sympleesmshn]

5. i think it is right

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

[s-cubed]

13. since it's abs of x, you get -1 for x < 0 and +1 for x>0, and

undefined at x = 0

[barackmyworld]

27. correct--here is how I remember the chain rule

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)

[LiberalBadger]

10. You got it...just simplify

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)

[Catch22Dem]

17. Hey welcome to DU

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

[LiberalBadger]

19. thanks! n/t

[gristy]

15. seems mostly right to me

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...

[JVS]

20. True calculus comes from pencil and paper, not from an infernal machine

[MikeDuffy]

21. dy/dx = -1 (x less than 0) , =+1 (x greater than 0) and undefined at x=0

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).

[s-cubed]

23. interesting set of posts

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.

[Liberal Classic]

28. As noted by others on this thread

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