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anonymous

  • 5 years ago

find y' assuming that the equation deermines a differentiable function f such that y =f(x) y^2+1=x^2 3y topic: implicit differentiation

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  1. anonymous
    • 5 years ago
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    y^2 +1 = x^2 sec y

  2. amistre64
    • 5 years ago
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    3y or secy?

  3. anonymous
    • 5 years ago
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    take the derivative of both sides wrt x

  4. anonymous
    • 5 years ago
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    2yy'=2x sec(y) +x^2 sec(y)tan(y)y'

  5. anonymous
    • 5 years ago
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    solve for y' using algebra

  6. anonymous
    • 5 years ago
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    amistre hello!

  7. anonymous
    • 5 years ago
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    i think it was \[y^2=x^2\sec(y)\] yes?

  8. amistre64
    • 5 years ago
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    howdy.... openstudys behaving bad tonight. i go to post and i get screenfreeze

  9. anonymous
    • 5 years ago
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    me too!

  10. amistre64
    • 5 years ago
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    2y y' = 2x3y + x^2 3 y' 2y y' -x^2 3 y' = 6xy y'(2y -3x^2) = 6xy y' = 6xy/(2y-3x^2)

  11. anonymous
    • 5 years ago
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    y^2 + 1 = x^2 sec y satellite

  12. anonymous
    • 5 years ago
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    so i used the product rule for the x^2 sec y part?

  13. anonymous
    • 5 years ago
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    thanks guys

  14. anonymous
    • 5 years ago
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    can you guys help me with the question i just posted

  15. anonymous
    • 5 years ago
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    yes the product rule. and when you take the derivative of sec(y) make sure to put the y' at the end. i will write it out if you like let me kn ow

  16. anonymous
    • 5 years ago
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    i can do it thank you

  17. anonymous
    • 5 years ago
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    2yy' = (2x)(secy)+(secy'tany')(x^2) is it correct?

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