Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

malexander

I am having trouble understanding the concept of implicit differentiation. Can anyone explain it in simple terms?

  • one year ago
  • one year ago

  • This Question is Closed
  1. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  2. malexander
    Best Response
    You've already chosen the best response.
    Medals 0

    can you give an example @lgbasallote ?

    • one year ago
  3. satellite73
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • one year ago
  4. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

    a good example would be \[y = a^x\] classic

    • one year ago
  5. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  6. malexander
    Best Response
    You've already chosen the best response.
    Medals 0

    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 :-(

    • one year ago
  7. malexander
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  8. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  9. satellite73
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  10. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  11. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  12. malexander
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  13. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  14. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  15. malexander
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  16. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  17. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

    now combine all the info

    • one year ago
  18. malexander
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  19. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  20. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  21. malexander
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

    • one year ago
  22. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

    y^2 stayed the same in 2xy^2

    • one year ago
  23. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

    in the second term x^2 remains the same

    • one year ago
  24. malexander
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  25. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

    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) ?"

    • one year ago
  26. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

    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

    • one year ago
  27. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  28. malexander
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • one year ago
  29. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  30. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  31. malexander
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  32. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  33. malexander
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  34. malexander
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  35. lgbasallote
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  36. malexander
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.