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Sqrt(2-5*x^2)

i did this: 2-5x^2) ^1/2

then 1/2(2-5x^2)

uh-oh got me there... may i ask how?

i think i kind of do but im not sure how to proceed it.

Confused? :O Too much stuff going on?

im a little confused with how you got to 10x

isn't it this: f(x)'*(g(x)+f(x)*g(x)' ?

i see but what happens to the half?

It's a tricky concept to get used to :(

Haha

I like that one :P

Uh oh, your brain esplode? :(
Want to see a nice easy example? One which includes no potatoes? :o

i understand the -10x but what happens to the -1/2 exponent.

lol

Next time we should do f ' (wal-mart)

We're making a copy of the `inside` and taking it's derivative. The -1/2 is on the `outside`.

We're not making a copy of the `outside`, if that makes sense :o

i know the chain rule but i dont understand the other one

after you find -10x then what

That last example was so ... unnecessary lol...
I should delete that.

Haha, we need the humor with these things sometimes, though : )

noooo dont lol

don't delete :) thank you

now, the critical points are= x=10?

i mean for x

i know how to solve for y

Since `multiplication` is commutative*
Now we need to find critical points?

I kinda threw a bunch of steps in there when I simplified.
Did any of that confuse you? :o

the chain rule, yes

oh simplied? uhhh no not really. i know how you did that.

yah that's the inner thing :o
take it's derivative, and multiply by it.

rewrite and first and multiply by the inner's derivative

okay i caught up with ya.

now to find that x.....

because they;re asking for the critical points.

I didn't want to interfere xD

it's okay, team work =]

x=5?

those are not critical points..... :/ it should be written like this.. ex: (0,0)

From that first zero? Hmm it looks like if we divide both sides by -5, we get x=0.

I'm not sure where you're getting 5 from :o

oops! right x=0

How bout this:
\[\large 0=\sqrt{2-5x^2}\]
What critical points is it giving us?

does not exist?

Squaring both sides gives us,\[\large 0=2-5x^2\]Solving for x gives us,\[\large x=\pm \sqrt{2/5}\]

wait a min i had that... :/ but it was x= sqrt(2/-5)

because the - is there^..

oops, right again.

@ zepdrix thank you so much for your time and effort. i appreciate lot's! goodnight.

np c: cya