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calyne Group Title

If f(x) + x^2 * f(x) = 10 and f(1)=2, find f'(1).

  • 2 years ago
  • 2 years ago

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  1. yash007 Group Title
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    check ur ques. again its wrong it does not satisfy for f(1)=2 in the main equation

    • 2 years ago
  2. calyne Group Title
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    well that's the friggen problem verbatim from the textbook

    • 2 years ago
  3. calyne Group Title
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    OH pellet

    • 2 years ago
  4. calyne Group Title
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    it's not sorry geez sorry it's f(x) + x^2 * f(x)^3 = 10 the second f(x) term is cubed

    • 2 years ago
  5. AnimalAin Group Title
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    Maybe rearrange the problem a bit....\[y+x^2y=10\implies y=\frac{10}{1+x^2}\]I don't think your problem meets the specified conditions.

    • 2 years ago
  6. yash007 Group Title
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    is ur anwer 2/5??

    • 2 years ago
  7. calyne Group Title
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    it's -16/13 that's the answer in the book

    • 2 years ago
  8. calyne Group Title
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    animalain - it's y + (x^2)(y^3) = 10

    • 2 years ago
  9. AnimalAin Group Title
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    Yeah, I got that. Trying to figure out my strategy....

    • 2 years ago
  10. yash007 Group Title
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    yeah its done

    • 2 years ago
  11. calyne Group Title
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    ...

    • 2 years ago
  12. AnimalAin Group Title
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    How did you do it?

    • 2 years ago
  13. yash007 Group Title
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    differentiate it n then put x=1 in equation u'll get the result

    • 2 years ago
  14. yash007 Group Title
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    is that done??

    • 2 years ago
  15. AnimalAin Group Title
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    \[y + (x^2)(y^3) = 10\implies y'+2xy^3+3x^2y^2y'=0\]\[\implies y'(1+3x^2y^2)=-2xy^3\implies y'=\frac{-2xy^3}{1+3x^2y^2}\]\[\implies y'(1)=\frac{-16}{13}\]Got it. Need to practice my calculus a little more....LOL

    • 2 years ago
  16. yash007 Group Title
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    all the best..:)

    • 2 years ago
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