Question

I've been learning derivatives for a week or so now, and have picked it up well, but this one problem is really sticking me. I know the chain rule and how to do higher order derivatives and such, but in piecing all the information together something seems to be getting lost. Anyone want to fill me in on the rules I might be missing or the mistakes that might be going on?
Here's what it says:
If x^2 + y^2 = 2y, show y'' = 1/((1-y)^3)
I continuously have problems with y disappearing from the equation, and I'm not totally sure what the rules are in using substitutions at different orders.

Answer 1

first solve for y what do you get

Answer 2

you need to use implicit differentiation here...

Answer 3

Can't really solve for y here, should do implicit differentiation.

Answer 4

familiar with that @katy ?

Answer 5

@Algebraic! *fist bump* I'll let you take this one.

Answer 6

Ugh! No. Crazy calc teacher. Last time we had a "review" it was supposed to cover the basics and I ended up having to read ahead in the book to teach myself the chain rule and some other stuff. I should've known it was the same deal here. Apparently it has to do with the different textbooks we have or something.
So, implicit differentiation, huh? Yeah, that means very little to me. What is this?

Answer 7

go to you tube and watch video its a great tool to use type in implicit differentiation

Answer 8

Implicit differentiation is really just chain rule.

Answer 9

The implication is that y is a function of x, so if you d/dx something like y^2, it is 2y*y' : where y' = dy/dx.

Answer 10

I always forget this thing... ugh.
Can you show me how you use this chain rule in implicit differentiation?\[\rm D\left[f(g(x))\right] =g'(x)f'(g(x)) \]

Answer 11

For this example,
x^2 + y^2 = 2y
differentiating term by term ->
2x +2yy' = 2y'

Answer 12

No, no, no... can you show that \(\rm {d \over dx}y^2 = 2yy'\) using chain rule?

Answer 13

ah, yes.
y^2 = (f(x))^2 =g(f(x)) : g(x) = x^2
d/dx (f(x))^2 = 2f(x)*f'(x).

Answer 14

Is that what you mean?

Answer 15

I don't really get it. :p

Answer 16

I see.

Answer 17

Well, @katy , do you see that d/dx (x^2 + y^2 = 2y) = 2x +2yy' = 2y' ?

Answer 18

I'm just reading through my textbook now; I'm not totally sure how the y is involved. So my understanding is that I'm going to differentiate it like I do the x, but then the primes come in and I get confused.

Answer 19

y' = dy/dx

Answer 20

So in order to differentiate ys like I do the xs, I have to multiply the differentiated value with y by the derivative with regards to x? So if there was x^5+4y^2=y^(1/2), dy/dx would be 5x^4+8yy'=(1/2)y^(-1/2)y'?

Answer 21

Yes.

Answer 22

Yay! I think I can get it from here. Three cheers for late night thinking, internet connections, and coffee! Those are what will be getting me through this class. Thanks for the help! I would've been sitting here for such a long time trying to do it without knowing that (and getting nothing done at all) if you guys hadn't chipped in when you did--thanks for being awesome!! God bless <3

Answer 23

Wonderful!
Take care,