If f(x)=8arcsin(x2), find the derivative. Help me please! Thanks so much!
integral number [18] in the back of my book says\[[18]\qquad\int\frac{\text d x}{\sqrt{a^2-x^2}}=\arcsin\left(\frac xa\right)+c\]
@donnac10 can you follow the formula?
Yeah.. I'm trying to see if I get the right answer. :)
Wait.. I'm kinda confuse now.. >_< Cause I see integral and I'm looking for a derivative.. >_<
( arcsin u ) = u' / sqrt ( 1 - u )
u = x^2 u' = ...
@donnac10 WHERE ARE YOU?
I'm sorry I was trying it out.. >_<
Can you take ( x^2) ' = ?
yeah.. It's 2x right?!?
Yup, pls just plug u and u' into the formula!
( arcsin u )' = u' / sqrt ( 1 - u )
@donnac10 Can you?
Yes, thanks so much for your help! Greatly appreciated! :D
@Chlorophyll I have another question. Sorry for asking too much.. >_< So what happens to the 8 in front of it?!?
It's the constant, keep it !
= 16x / sqrt ( 1 - x^2 )
Cause that's what I put on this homework website that I have been working on and it says it's wrong.. -_-
Put it here!
Which one?!?
The one you say wrong, so I can see why
Isn't it the 16x / sqrt ( 1 - x^2 )?!?
Try it: 16x / sqrt ( 1 - x^4 )
It's right! How did you do that?!?
I'm sorry that I typo in the formula ( arcsin u ) = u' / sqrt ( 1 - u ^2 )
u = x^2 => u^2 = (x^2) ^2 = x^4
That's what I though.. I missed that! And I have been doing it for days.. Haha! Thanks so much for pointing that out! You're very helpful.. :D
That's the whole reason we need the asker show up :)
Pls, not to be annoyed if I keep calling out your name ;)
Thanks so much! Oh no.. It's actually quite helpful because I actually see that I have a notification.. Cause when people just reply I don't really see it.. >_< Thanks for being so patient!
That's how annoyed aggressive I am. Thanks for keep up with my temper ;)
Ohh really?!? It's cool... I think I work better when people are like that to me.. :P I'm like that too when I teach people.. So I totally understand.. :D
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