## malexander Group Title I am having trouble understanding the concept of implicit differentiation. Can anyone explain it in simple terms? 2 years ago 2 years ago

1. lgbasallote

1) differentiate left side 2) differentiate right side 3) isolate y' 4) substitute y (if possible)

2. malexander

can you give an example @lgbasallote ?

3. satellite73

simplest explanation i have ever seen is in the video here http://patrickjmt.com/ i cannot link directly to the video, but it is in the calculus section and very easy to understand

4. lgbasallote

a good example would be $y = a^x$ classic

5. lgbasallote

note: a is a constant and x is the variable you wanna differentiate

6. malexander

i saw that same video @satellite73 . it was okay-i tied to apply the same thing to a problem my professor gave me and i totally stalled :-(

7. malexander

for example: ysinx+x^2y^2=2x

8. lgbasallote

first term -> use product rule second term -> use product rule derivative of 2x would be 2

9. satellite73

think "$$y=f(x), y'=f'(x)$$ and use the product and chain rules where needed

10. lgbasallote

product rule on ysin x would be... $y(\cos x) + y' \sin x$ try product rule on second term

11. lgbasallote

when i said second term i meant X^2 y^2

12. malexander

so..... dy/dx sinx+ycosx+2x2y=2?

13. lgbasallote

not exactly....you use power rule on x^2 y^2

14. lgbasallote

*facepalm* product rule...darn these p's

15. malexander

so the product rule on x^2y^2 would be 2xy^2+x^2(dy/dx) ?

16. lgbasallote

not exactly...second term would be x^2 (2y)(dy/dx)

17. lgbasallote

now combine all the info

18. malexander

wait...how is y^2 becoming 2y if your doing the product rule (f'g+fg')...

19. lgbasallote

second term would be x^2 multipled to derivative of y^2 right?

20. lgbasallote

derivative of y^2 is 2y (dy/dx)

21. malexander

what im saying is that, if g(x) is y^2, the product rule calls for one of the g(x) not to become the g'(x)...so wouldnt one of the y^2 stay the same?

22. lgbasallote

y^2 stayed the same in 2xy^2

23. lgbasallote

in the second term x^2 remains the same

24. malexander

im still not getting the concept (im a bit slow tonight, sorry)

25. lgbasallote

by "second term" i am following your answer you said "so the product rule on x^2y^2 would be 2xy^2+x^2(dy/dx) ?"

26. lgbasallote

2xy^2 is correct...you held y^2 constant then took the derivative of x^2 for the second term you held x^2 constant but took the derivative of y^2 wrong

27. lgbasallote

derivative of y^2 is not simply dy/dx...it's 2y(dy/dx)

28. malexander

ohhhh okay! now i understand...so anytime you take the derivative of y...its not only its derivative but you put dy/dx at the end of it

29. lgbasallote

that's pretty much it....so now combine all the info

30. lgbasallote

i mean add the derivative of ysinx and the derivative of x^2y^2

31. malexander

so its: dy/dx sinx+ycosx+2xy^2+x^2 2y(dy/dx)=2

32. lgbasallote

yep you got it...now bring to the right side all the terms that doesnt include dy/dx

33. malexander

sooo dy/dx sinx+x^2 2y(dy/dx)=2-ycosx-2xy^2 so far

34. malexander

so is it: dy/dx=2-ycosx-2xy^2/sinx+x^2 2y ?

35. lgbasallote

yep...i dont suppose you can put y in terms of x so i guess that's the final answer...can you confirm @satellite73 ?

36. malexander

NICEEE :-D I never got this concept until now. Thanks so much for helping me out with this @lgbasallote . I will not close the question until @satellite73 responds to your query.