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anonymous

  • 4 years ago

Implicit differentiation: y + 3xy -4 = 0

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  1. anonymous
    • 4 years ago
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    i see you lol

  2. dumbcow
    • 4 years ago
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    take derivative of each term as you would normally, except whenever you differentiate y multiply by y'

  3. lalaly
    • 4 years ago
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    y'+3y+3xy'=0 y'+3xy'=4 y'(1+3x)=4 y'=4/(1+3x)

  4. dumbcow
    • 4 years ago
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    y'+3y+3xy'=0 y'+3xy'=4 <-- **4 should be -3y ** y'(1+3x)=4 <-- y'=4/(1+3x) <--

  5. anonymous
    • 4 years ago
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    Thank you, that is what my book says as well. I'm a little bit lost though. I get that the derivative of the y = 1, multiply that by y\[\prime\], but then 3xy, why does that become 3xy[\prime\]? And the -4, how does that become 3y[\prime\]?

  6. anonymous
    • 4 years ago
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    Oh, am I to treat it as (3xy)\[^{1}\]?

  7. anonymous
    • 4 years ago
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    Ah sorry, my equation formatting is a mess :\

  8. dumbcow
    • 4 years ago
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    its the product rule (fg)' = fg' +f'g

  9. dumbcow
    • 4 years ago
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    (3xy)' = (3x)*y' + (3x)'*y = 3xy' + 3y

  10. anonymous
    • 4 years ago
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    I see. So then the 4 still becomes zero, as in other cases?

  11. dumbcow
    • 4 years ago
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    right

  12. anonymous
    • 4 years ago
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    Okay, thank you very much :D That will be very helpful for the next problems, hopefully I can manage them on my own.

  13. dumbcow
    • 4 years ago
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    yw :)

  14. anonymous
    • 4 years ago
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    Just double checking before I continue, for example: 3*x^3*y^2 would become 3x^3(2y) + 3(3x^2)(y^2), correct?

  15. dumbcow
    • 4 years ago
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    yes but don;t forget about the y'

  16. anonymous
    • 4 years ago
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    So, each of those two terms is also multiplied by y prime? I'm not to clear on what you mean.

  17. dumbcow
    • 4 years ago
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    no only the term where you differentiated y (1st term) 3x^3(2y)y' because 2y came from differentiating y^2

  18. anonymous
    • 4 years ago
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    Ah okay. So, I multiply by y(prime) whenever I differentiate y.

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