## Ika_Kim 2 years ago ive tried to prove it many times but , i still didnt get it.... if dy/dx = -y/x+2y prove that d''y/dx''=8/(x+2y)^3

1. tkhunny

Do you mean $$\dfrac{-y}{x+2y}$$? That is NOT what you have written. Folks in calculus should know about the Order of Operations. Have you considered the Quotient Rule for differentiation?

2. precal

use the quotient rule

3. Ika_Kim

really ? sorry i dont know how to write it in the right way...yes i use that quotient rule , but then i got this$\frac{ -\frac{ dy }{ dx }.x+2y-(2y.\frac{ dy }{ dx }) }{ (x+2y)^2 }$ then what should i do ?

4. tkhunny

You should try that again. The whole numerator is a mess.

5. Ika_Kim

ok..wait

6. Ika_Kim

someby pls help me..i dont get it..... :{

7. tkhunny

Numerator: (x+2y)(-y') - (-y)(1+2y') Look at it very carefully.

8. Ika_Kim

oh yeahh my numerator is really a mess..then what should i do ?

9. Ika_Kim

what should i do ~~~~~ omgg

10. hartnn

even i tried this more than once, and i don't get numerator = 8, is anything else given ?

11. Ika_Kim

the original question .

12. Ika_Kim

idk what im doing -.-

13. Ika_Kim

first question

14. hartnn

ohhh...so, xy+y^2 =4.... then its easy, did you get that numerator first ? next thing will be to simplify it.

15. hartnn

got this : (x+2y)(-y') - (-y)(1+2y') first ?

16. Ika_Kim

yes i got it just now .haha

17. Ika_Kim

then ?

18. hartnn

now use, y'=-y/(x+2y)

19. Ika_Kim

yes im using that..then quotient rule ?

20. hartnn

(x+2y)(-y') - (-y)(1+2y') is the result of numerator AFTER applying quotient rule, now you have to simplify

21. tkhunny

Just for the record, it is ALWAYS more beneficial to show the ORIGINAL problem statement.

22. hartnn

yeah, that would have saved me few minutes...

23. Ika_Kim

to be honest im stucking again

24. hartnn

what u got after substituting y'=-y/(x+2y) in that numerator ?

25. Ika_Kim

$2y-\frac{ 2y^2 }{ x+2y }$ what about this ? for that numerator

26. Ika_Kim

$\frac{ 2xy+2y }{ (x+2y)^3 }$

27. Ika_Kim

this ???????

28. Ika_Kim

then subtitute (0,2) into that equation ?

29. hartnn

its actually this : $$\huge \frac{ 2xy+2y^2 }{ (x+2y)^3 }=\frac{ 2(xy+y^2) }{ (x+2y)^3 }$$ now use xy+y^2 =4 and you are done

30. hartnn

$$2y-\frac{ 2y^2 }{ x+2y }$$ was correct

31. Ika_Kim

$\frac{ 2xy+2y^2}{(x+2y)^3}$

32. Ika_Kim

this ????am i right ??

33. hartnn

yeah, factoring out 2 will give you 2 (xy+y^2) and xy+y^2 as per the question is 4

34. Ika_Kim

ohhhh........i see ur answer........and i totally got it ! thanks a lot @hartnn !!:) really appreciate ^_^

35. hartnn

welcome ^_^