## 2pie Group Title find (dy/dx) by implicit differentiation (x/y)=(y/x) 11 months ago 11 months ago

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1. iambatman

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2. iambatman

Quotient rule on both side

3. 2pie

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4. 2pie

is it correct? what do I do next

5. 2pie

@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. 2pie

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. 2pie

yes

14. 2pie

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. 2pie

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

17. 2pie

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. 2pie

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

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. 2pie

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. 2pie

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. 2pie

yes by dividing 2y on both sides

32. hartnn

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

33. 2pie

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. 2pie

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. 2pie

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. 2pie

thanks ;)