need some math help

[platinumike]

Its actually for calculus and were doing derivatives. The original problem:

y=x^2 (squareroot)1-x^2

which is read as: y equals x squared times the square root of 1- x squared

I rewrite it, do the product rule for derivatives and end up with this:

-x^3(1-x^2)^-1/2 + 2x(1-x^2)^-1/2

Iam stuck at this point. Apparently the book is factoring out one of thoose quantities raised to the -1/2 and getting this answer:

x(1-x^2)^-1/2 ( (-x^2(1)+2(1-x^2) )

I feel like im gonna get no replies the way i typed this out, worth a try though! apprarently its factoring that i didnt learn in algebra.

 

[AccruedExpenditure]

The answer is: do your own homework

 

[platinumike]

is it a crime to ask for help?

 

[QED]

Originally posted by: platinumike
is it a crime to ask for help?

To people who don't know they answer it is.

To me, I don't mind helping someone if they've shown they put the effort into it.

 

[Koing]

My bro would probably p!ss all over that...but alas I don't have ihs Maths skills so I can't help you out

Other guys, give the guy some help if you can. No point in telling him to do his own homework...

Koing

 

[platinumike]

I've put the effort into this. I've been working on this for the past 30 minutes. You honestly think I would put no effort into this problem and type it out here on atot and waitfor a response? that actually seems easier? I type it here because I have no clue and you guys are great at math.

 

[dullard]

Just factor it out yourself. You'll get the book's answer.

Edit: Wait, you have a minus sign error.

 

[Vegitto]

I can do this, I know I can, but what is square root? Is it the same as root? Like:

Square root of 16 = 4, because 4^2 = 16?

If so, I'll do this .

 

[QED]

Ok... you need to first you the product rule, and then the chain rule...

hold on... I'll elaborate...

 

[platinumike]

thank you. Yes I used the product rule and the chain rule, do a little bit of simplifying and I get stuck at that point.

EDIT: had to edit what i said here before, i was wrong.

 

[dullard]

If the math looks awful, and you think you can't do it, then simplify. Let z = (1-x^2)^0.5. Then the problem becomes this:

y = x^2 * z

Take the derivative: y' = 2x*z + x^2*z' = 2x*z - x^3/z. Note your minus sign error here in the exponent.

Simplify: y' = x*(2z-x^2/z) = x/z*(2z^2-x^2) = matches the book.

 

[QED]

Let f(x) = x^2, g(x) = (1-x^2)^.5. Then y = f(x) * g(x)

By the product rule:

y' = f'(x) * g(x) + g'(x) * f(x).

Now f'(x) = 2x.

To calculate g'(x), we now use the chain rule:

Let m(x) = 1 - x^2, n(x) = x^.5 . Thus, g(x) = m( n(x) ). By the chain rule,
g'(x) = m'( n(x) ) * n'(x). The deriviate of m and n are
m'(x) = -2x and n'(x) = .5 * x^(-.5). Hence

g'(x) = -2( .5 * x^(-.5)) = - x^(-.5).

So,

y' = 2x * (1 - x^2)^.5 + x^2 * (- x^(-.5))
= 2x * (1 - x^2)^.5 - x^(1.5)


You can simplify from there...

 

[Vegitto]

You guys are weird. I think my solution is much better, although unfinished. (I'm only 15, though, and this should exceed any knowledge I have .)

y = (x^2)(sqrt(1-(x^2)))
y^2 = (x^4)(1-(x^2))
y^2 = x^4 - x^6
y = sqrt(x^4 - x^6)

My TI-83 thinks it's correct, but I can't simplify it any further .
Can you take it from here?

 

[chuckywang]

Originally posted by: Vegitto
You guys are weird. I think my solution is much better, although unfinished. (I'm only 15, though, and this should exceed any knowledge I have .)

y = (x^2)(sqrt(1-(x^2)))
y^2 = (x^4)(1-(x^2))
y^2 = x^4 - x^6
y = sqrt(x^4 - x^6)

My TI-83 thinks it's correct, but I can't simplify it any further .
Can you take it from here?

Great, so you didn't take a single derivative and we are, like Maxine Nightingale says, right back where we started from.

 

[Vegitto]

Originally posted by: chuckywang

Originally posted by: Vegitto
You guys are weird. I think my solution is much better, although unfinished. (I'm only 15, though, and this should exceed any knowledge I have .)

y = (x^2)(sqrt(1-(x^2)))
y^2 = (x^4)(1-(x^2))
y^2 = x^4 - x^6
y = sqrt(x^4 - x^6)

My TI-83 thinks it's correct, but I can't simplify it any further .
Can you take it from here?

Great, so you didn't take a single derivative and we are, like Maxine Nightingale says, right back where we started from.

So x^4 - x^6 can't be simplified? My math world is collapsing!

 

[dullard]

Originally posted by: Vegitto
You guys are weird. I think my solution is much better, although unfinished. (I'm only 15, though, and this should exceed any knowledge I have .)

It does exceed the knowledge you have. He is NOT solving for y. He is solving for the slope of the y vs x graph for ALL positions of x.

 

[chuckywang]

Originally posted by: Vegitto

Originally posted by: chuckywang

Originally posted by: Vegitto
You guys are weird. I think my solution is much better, although unfinished. (I'm only 15, though, and this should exceed any knowledge I have .)

y = (x^2)(sqrt(1-(x^2)))
y^2 = (x^4)(1-(x^2))
y^2 = x^4 - x^6
y = sqrt(x^4 - x^6)

My TI-83 thinks it's correct, but I can't simplify it any further .
Can you take it from here?

Great, so you didn't take a single derivative and we are, like Maxine Nightingale says, right back where we started from.

So x^4 - x^6 can't be simplified? My math world is collapsing!

*repeatedly pounds head into keyboard*

 

[Vegitto]

Originally posted by: dullard

Originally posted by: Vegitto
You guys are weird. I think my solution is much better, although unfinished. (I'm only 15, though, and this should exceed any knowledge I have .)

It does exceed the knowledge you have. He is NOT solving for y. He is solving for the slope of the y vs x graph for ALL positions of x.

Oh, he wants x?
Easy.

x^4 - x^6 = 0
x^4 = x^6
x = 0

 

[dullard]

Originally posted by: Vegitto
Oh, he wants x?
Easy.

x^4 - x^6 = 0
x^4 = x^6
x = 0

No. No, he does NOT want x. He does not want y or x.

 

[Vegitto]

Originally posted by: chuckywang

Originally posted by: Vegitto

Originally posted by: chuckywang

Originally posted by: Vegitto
You guys are weird. I think my solution is much better, although unfinished. (I'm only 15, though, and this should exceed any knowledge I have .)

y = (x^2)(sqrt(1-(x^2)))
y^2 = (x^4)(1-(x^2))
y^2 = x^4 - x^6
y = sqrt(x^4 - x^6)

My TI-83 thinks it's correct, but I can't simplify it any further .
Can you take it from here?

Great, so you didn't take a single derivative and we are, like Maxine Nightingale says, right back where we started from.

So x^4 - x^6 can't be simplified? My math world is collapsing!

*repeatedly pounds head into keyboard*

Sorry dude, I just tried to help..

 

[Vegitto]

Originally posted by: dullard

Originally posted by: Vegitto
Oh, he wants x?
Easy.

x^4 - x^6 = 0
x^4 = x^6
x = 0

No. No, he does NOT want x. He does not want y or x.

Then what does he want?

 

[chuckywang]

Originally posted by: Vegitto

Originally posted by: chuckywang

Originally posted by: Vegitto

Originally posted by: chuckywang

Originally posted by: Vegitto
You guys are weird. I think my solution is much better, although unfinished. (I'm only 15, though, and this should exceed any knowledge I have .)

y = (x^2)(sqrt(1-(x^2)))
y^2 = (x^4)(1-(x^2))
y^2 = x^4 - x^6
y = sqrt(x^4 - x^6)

My TI-83 thinks it's correct, but I can't simplify it any further .
Can you take it from here?

Great, so you didn't take a single derivative and we are, like Maxine Nightingale says, right back where we started from.

So x^4 - x^6 can't be simplified? My math world is collapsing!

*repeatedly pounds head into keyboard*

Sorry dude, I just tried to help..

He wants dy/dx, ie the derivative of y with respect to x, ie y'. He does not want to solve for y or whatever you were trying to do.

 

[dullard]

Originally posted by: Vegitto
Then what does he want?

From the original post: "and were doing derivatives".

From my first response to you: "slope of the y vs x graph".

Take your pick. They mean the same thing.

 

[Vegitto]

Originally posted by: dullard

Originally posted by: Vegitto
Then what does he want?

From the original post: "and were doing derivatives".

From my first response to you: "slope of the y vs x graph".

Take your pick. They mean the same thing.

Oh, wait! I know. I don't know what it's called in English, though .. And I don't think I can give the answer if the only possible outcomes are x=-1 through x=1 and y=0 consistently..

 

[MrScott81]

Originally posted by: Vegitto

Originally posted by: chuckywang

Originally posted by: Vegitto

Originally posted by: chuckywang

Originally posted by: Vegitto
You guys are weird. I think my solution is much better, although unfinished. (I'm only 15, though, and this should exceed any knowledge I have .)

y = (x^2)(sqrt(1-(x^2)))
y^2 = (x^4)(1-(x^2))
y^2 = x^4 - x^6
y = sqrt(x^4 - x^6)

My TI-83 thinks it's correct, but I can't simplify it any further .
Can you take it from here?

That's nice to try to help, but you shouldn't help if you have no idea what you're talking about.

Great, so you didn't take a single derivative and we are, like Maxine Nightingale says, right back where we started from.

So x^4 - x^6 can't be simplified? My math world is collapsing!

*repeatedly pounds head into keyboard*

Sorry dude, I just tried to help..