- #1
Hollysmoke
- 185
- 0
I just wanted to make sure I differentiated this correctly before I started graphing it:
y=x^2/1-x^2
y' = (2x)/(1-x^2)^2
y''=(2+6x^2)/(1-x^2)^3
y=x^2/1-x^2
y' = (2x)/(1-x^2)^2
y''=(2+6x^2)/(1-x^2)^3