Question

find d^2y/dx^2 in terms of x and y.
x^2-y^2=16

Answer 1

this is a calculus problem

Answer 2

You can use implicit differentiation..can you do that?

Answer 3

yes

Answer 4

thats what i"m trying to do but I don't get the right answer -16/y^3

Answer 5

Nice so we will first do d/dx for the whole function to get:
2x-2y*(dy/dx)=0
Are you okay with this first one?

Answer 6

yes

Answer 7

Then we will do the second d/dx for the whole function to get:
2-(2\frac{dy}{dx}+2y*\frac{d2y}{dx2})=0

Answer 8

Sorry, that didn't come out right...

Answer 9

You'll have:
\[2-(2\frac{dy}{dx}\frac{dy}{dx}+2y\frac{d2y}{dx2})\]

Answer 10

And that =0 of course.

Answer 11

why do you take the 2 out?

Answer 12

I had this:
2x-2y*(dy/dx)=0
Then d/dx of 2x is 2, and the second part is treated as one block covered by that negative sign.

Answer 13

Or are you asking about the 2 within the parentheses?

Answer 14

no I got it 2 is the derivative of 2x

Answer 15

Good. Can you take it from there?

Answer 16

no I can't

Answer 17

are you suppose to cancel out the derivatives?

Answer 18

No, what you want to do is to use algebra to make d2y/dx2 the subject of the equation. Leave the derivatives intact.

Answer 19

That should give you
\[\frac{d2y}{dx2}=\frac{2-2\frac{dy}{dx}\frac{dy}{dx}}{2y} \]

Answer 20

I have to be honest, I'm not entirely sure whether we should leave the \[\frac{dy}{dx}\frac{dy}{dx} \ \] in that form.

Answer 21

I think its just dy/dx

Answer 22

I'm more inclined to leave it as (dy/dx)^2 but you can confirm from other openstudy users. What do you think, @radar ?

Answer 23

Sorry but I can not offer any clarification. Hopefully another OC user can, or repost the question.