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malexander

  • 2 years ago

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

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  1. lgbasallote
    • 2 years ago
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    1) differentiate left side 2) differentiate right side 3) isolate y' 4) substitute y (if possible)

  2. malexander
    • 2 years ago
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    can you give an example @lgbasallote ?

  3. satellite73
    • 2 years ago
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    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
    • 2 years ago
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    a good example would be \[y = a^x\] classic

  5. lgbasallote
    • 2 years ago
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    note: a is a constant and x is the variable you wanna differentiate

  6. malexander
    • 2 years ago
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    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
    • 2 years ago
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    for example: ysinx+x^2y^2=2x

  8. lgbasallote
    • 2 years ago
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    first term -> use product rule second term -> use product rule derivative of 2x would be 2

  9. satellite73
    • 2 years ago
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    think "\(y=f(x), y'=f'(x)\) and use the product and chain rules where needed

  10. lgbasallote
    • 2 years ago
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    product rule on ysin x would be... \[y(\cos x) + y' \sin x\] try product rule on second term

  11. lgbasallote
    • 2 years ago
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    when i said second term i meant X^2 y^2

  12. malexander
    • 2 years ago
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    so..... dy/dx sinx+ycosx+2x2y=2?

  13. lgbasallote
    • 2 years ago
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    not exactly....you use power rule on x^2 y^2

  14. lgbasallote
    • 2 years ago
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    *facepalm* product rule...darn these p's

  15. malexander
    • 2 years ago
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    so the product rule on x^2y^2 would be 2xy^2+x^2(dy/dx) ?

  16. lgbasallote
    • 2 years ago
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    not exactly...second term would be x^2 (2y)(dy/dx)

  17. lgbasallote
    • 2 years ago
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    now combine all the info

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

  19. lgbasallote
    • 2 years ago
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    second term would be x^2 multipled to derivative of y^2 right?

  20. lgbasallote
    • 2 years ago
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    derivative of y^2 is 2y (dy/dx)

  21. malexander
    • 2 years ago
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    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
    • 2 years ago
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    y^2 stayed the same in 2xy^2

  23. lgbasallote
    • 2 years ago
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    in the second term x^2 remains the same

  24. malexander
    • 2 years ago
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    im still not getting the concept (im a bit slow tonight, sorry)

  25. lgbasallote
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    derivative of y^2 is not simply dy/dx...it's 2y(dy/dx)

  28. malexander
    • 2 years ago
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    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
    • 2 years ago
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    that's pretty much it....so now combine all the info

  30. lgbasallote
    • 2 years ago
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    i mean add the derivative of ysinx and the derivative of x^2y^2

  31. malexander
    • 2 years ago
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    so its: dy/dx sinx+ycosx+2xy^2+x^2 2y(dy/dx)=2

  32. lgbasallote
    • 2 years ago
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    yep you got it...now bring to the right side all the terms that doesnt include dy/dx

  33. malexander
    • 2 years ago
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    sooo dy/dx sinx+x^2 2y(dy/dx)=2-ycosx-2xy^2 so far

  34. malexander
    • 2 years ago
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    so is it: dy/dx=2-ycosx-2xy^2/sinx+x^2 2y ?

  35. lgbasallote
    • 2 years ago
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    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
    • 2 years ago
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    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.

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