## anonymous 3 years ago find (dy/dx) by implicit differentiation (x/y)=(y/x)

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1. anonymous

|dw:1381786279195:dw|

2. anonymous

Quotient rule on both side

3. anonymous

|dw:1381786349930:dw|

4. anonymous

is it correct? what do I do next

5. anonymous

@iambatman

6. myininaya

Good so far.

7. myininaya

$\frac{y-xy'}{y^2}=\frac{xy'-y}{x^2}$ I would clear the fractions first.

8. myininaya

Multiply both sides by x^2y^2.

9. myininaya

$x^2y^2 \frac{y-xy'}{y^2}=x^2y^2\frac{xy'-y}{x^2}$ You see that we can write this without the fractional style now, right?

10. anonymous

yes by calcelling the $y ^{2} x^{2}$ on each side respectively $x^{2}(y-xy^1)+y^2(xy^1-y)$ then?

11. myininaya

Distribute. Put y' terms on one side and the other non-y' terms on the other side Then factor y' out. The divide by whatever is being multiplied by y'.

12. myininaya

You mean equal right not plus?

13. anonymous

yes

14. anonymous

After I multiply I got$x^2y-x^3y^\prime =xy^2y^\prime-y^3$

15. myininaya

Right. Continue. Put y' terms together on one side and then non-y' terms on the other.

16. anonymous

$x^2y+y^3=xy^2y^\prime + x^3 y^\prime$

17. anonymous

then im not sure what to do

18. myininaya

Factor the y' out on that one side.

19. myininaya

$x^2y+y^3=y'(xy^2+x^3)$

20. myininaya

Divide both sides by what y' is being multiplied by.

21. myininaya

22. anonymous

$x^2y+y^3/xy^2+x^3$=$y^\prime$

23. myininaya

$y'=\frac{x^2y+y^3}{xy^2+x^3} =\frac{y(x^2+y^2)}{x(y^2+x^2)}$ Anything cancel?

24. myininaya

Now instead of starting the way you guys did at the beginning. I think @hartnn would like to show you a easier way.

25. anonymous

ok show me

26. hartnn

hey 2pie since we have x/y = y/x can we say x^2 = y^2 ?

27. hartnn

i just did a cross-multiplication

28. anonymous

yes

29. myininaya

I think the slope can be simplified even more by the way when you guys are done. :)

30. hartnn

and then we can differentiate x^2=y^2 to get 2x =2y dy/dx can you isolaye dy/dx from this ?

31. anonymous

yes by dividing 2y on both sides

32. hartnn

and what would u get finally ? as dy/dx ?

33. anonymous

dy/dx = x/y

34. myininaya

And recall from your initial equation y/x=x/y

35. myininaya

Now another way... x^2=y^2 or x^2-y^2=0 or (x-y)(x+y)=0 or x-y=0 or x+y=0 or y=x or y=-x If y=x, then y'=1 If y=-x, then y'=-1

36. anonymous

wow it was easier than I thought thank you very much for your help and time @myininaya @hartnn :)

37. hartnn

i see you are new here , so $\begin{array}l\color{red}{\text{W}}\color{orange}{\text{e}}\color{#e6e600}{\text{l}}\color{green}{\text{c}}\color{blue}{\text{o}}\color{purple}{\text{m}}\color{purple}{\text{e}}\color{red}{\text{ }}\color{orange}{\text{t}}\color{#e6e600}{\text{o}}\color{green}{\text{ }}\color{blue}{\text{O}}\color{purple}{\text{p}}\color{purple}{\text{e}}\color{red}{\text{n}}\color{orange}{\text{S}}\color{#e6e600}{\text{t}}\color{green}{\text{u}}\color{blue}{\text{d}}\color{purple}{\text{y}}\color{purple}{\text{!}}\color{red}{\text{!}}\color{orange}{\text{ }}\color{#e6e600}{\text{:}}\color{green}{\text{)}}\color{blue}{\text{}}\end{array}$

38. anonymous

thank you Im really glad I joined

39. myininaya

Who couldn't pass up free math help?

40. hartnn

we too are glad that you joined :)

41. anonymous

thanks ;)