## mathcalculus one year ago Help:implicit differentiation with a fractions :(

1. mathcalculus

2. myko

multiplying everything by 36 you get; 2x^2+y^2=36 which maybe more clear for you, :)

3. mathcalculus

hey! :) i tried this: x²/16 + y²/36 = 1 (1/16)x² + (1/36)y² = 1 2(1/16)x + 2(1/36)yy' = 0

4. mathcalculus

but so far im not sure what to do after because i know i get x/8 + ?? but then i can't really multiply (1/36) 2y*dy/dx ?

5. mathcalculus

isn't the point here to find dy/dx?

6. myko

let \(F(x,y)=2x^2+y^2=36\) then: \(dF(x,y)=4xdx+2ydy=0\) from here: \(y'=dy/dx=-4x/2y=-2x/y\)

7. mathcalculus

huh? o.o

8. mathcalculus

howd you get this?? let F(x,y)=2x2+y2=36

9. myko

\(dF(x,y)\) is colled full differential it is: \(dF(x,y)=F_xdx+F_ydy\)

10. mathcalculus

i never seen that before

11. myko

\(F_x\) is partial derivative respect to x

12. mathcalculus

ok

13. mathcalculus

i know those, but where are the numbers coming from?

14. mathcalculus

in what order, where did you take them?

15. mathcalculus

i know they're given but how did you get 36 on the other side. and 2x^2+y^2let F(x,y)=2x2+y2=36

16. myko

ok, other way: think that y is actualy a function of x, so it is y=y(x). Then: \(2x^2+y(x)^2=36\) now we differentiate keeping this in mind 4x+2yy'=0 now just solve for y'

17. mathcalculus

2x2+y(x)2=36?

18. mathcalculus

yes i know when to differentiate but how did you change the function to that

19. myko

Ya you right I made a mistake. I was treating 36 like 32, :). Shold be. Rest of steps same. dw:1381788857534:dw|

20. myko

|dw:1381789088155:dw|

21. mathcalculus

i know I was going to tell you.

22. mathcalculus

you can't have 36 at the end.

23. mathcalculus

i saw a different way on yahoo, would be be able to help me with this one x²/16 + y²/36 = 1 (1/16)x² + (1/36)y² = 1 2(1/16)x + 2(1/36)yy' = 0 (x/8) + (y/18)y' = 0 (y/18)y' = -(x/8) y' = -(x/8)/(y/18) y' = -(x/8)(18/y) y' = -(9x/4y) y'(2) = -(9(2))/(4(5.2)) y'(2) = -18/20.8 y'(2) = -9/10.4 y'(2) = -0.865

24. mathcalculus

I'm not really sure how they got the 4th step :/.... this part----> + (y/18)*dy/dx ??

25. myko

2(1/16)x + 2(1/36)yy' = 0 (2/16)x+(2/36)yy'=0 (x/8) + (y/18)y' = 0

26. mathcalculus

yes my question is exactly what you wrote up there .... |dw:1381789520628:dw|(x/8) + (y/18)y' = 0

27. myko

(2/36)=1/18

28. mathcalculus

how did it get simplified to y/18*dy/dx?

29. myko

your answer is same, just simplify 2 in numerator with 36 in denominator

30. mathcalculus

thanks! :) i dontkknow why i missed that silly stuff.

31. myko

:)

32. mathcalculus

then i plug in for x and y and find the common denominators rights?

33. myko

yes

34. mathcalculus

wait we're not done yet! :(

35. mathcalculus

i got -.005379

36. mathcalculus

the question is asking y'(1)

37. myko

y' = -(9x/4y) so: substitute x=1, y=5.80948 y'(1)=-9/4(5.80948)

38. mathcalculus

okay i got dy/dx=-.387298

39. mathcalculus

got it thanks @myko :))!

40. myko

yw