davidiswhat
New member
- Joined
- Oct 21, 2014
- Messages
- 6
dy/dx = (1-y^2)^1/2
find d^2y/dx^2
A: -y
so to get it ...
1/2(1-y^2)^-1/2(-2y) //chain rule ... dont get why a dy/dx should be on top
-2y/2dy/dx
So to get the answer im suppose to cancel the denominator by dy/dx and i wish to know how/why we got that. i kinda get that we're solving y'(y) instead of y'(x) but im sill confuse. Also i think we used to do something like dx/dx when we did y'(x) so if it was a x instead ... y''(x)=1/2(1-x^2)^-1/2(-2x) dx/dx why is this here?
basically
-y(dy/dx) <- ?
/dy/dx