## anonymous 5 years ago compute d^2/dx^2 at the point (4,2) x^2+y^2=20

1. amistre64

is that our f(x) sitting there?

2. amistre64

2x x' + 2y y' = 0 2x + 2y y' = 0 y' = -x/y

3. amistre64

y'' = -1/y + y'x/y^2

4. amistre64

$y'' = -\frac{1}{y} + \frac{{-x^2}}{y^3}$

5. amistre64

(x=4, y=2) y'' = -1/2 - 16/8

6. amistre64

y'' = -32/8 = -4 perhaps?

7. amistre64

thats wrong

8. anonymous

why not 5/2?

9. amistre64

-4/8 - 16/8 = -20/8 y'' = -5/2

10. amistre64

i was gettin there, my head on holds so much at once lol

11. anonymous

cool, thanks!

12. amistre64

i was wondering if i could have done that from this step like this: 2x + 2y y' = 0 2 + 2y' + 2y y'' = 0 y'' = -2 -2(-x/y) -1 -(-x/y) -1 -(-2) ---------- = -------- = ------- = doesnt look like it 2y y 2

13. anonymous

that looks more complicated, do you happen to also know about definite integrals?

14. amistre64

i can do them with somewhat of skill and mostly luck ;)

15. amistre64

definite integrals are integrating within an interval

16. anonymous

lol okay, let's see how lucky you are, I'm about to post one =D

17. amistre64

yay!!

18. anonymous

okay, for some reason it won't let me ask a question like normal

19. amistre64