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anonymous

  • 5 years ago

find the slope of the tangent line to the curve -1x^2-2xy-3xy^3=-185 at point (-1,4)

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  1. anonymous
    • 5 years ago
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    sorry i mean at point (-2,-4)

  2. anonymous
    • 5 years ago
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    The derivative if this function is-2x-2y-2xy'-3y^3-3x3y^2y'

  3. anonymous
    • 5 years ago
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    Y'=...... Make the (-2,-4)in

  4. anonymous
    • 5 years ago
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    oh crap i wrote the equation wrong... sorry it's actually -1x^2-4xy-2y^3=92

  5. anonymous
    • 5 years ago
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    The derivative if this function is-2x-2y-2xy'-3y^3-3x3y^2y'=-185.so y'=. And then make (-2,-4)in.

  6. anonymous
    • 5 years ago
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    ok so if we're taking the derivative of this equation -1x^2-4xy-2y^3=92

  7. anonymous
    • 5 years ago
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    would it be -2x-4xy'=4y-6y^2y'=0

  8. anonymous
    • 5 years ago
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    No, it is a implicit function.

  9. anonymous
    • 5 years ago
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    yes it is, i took the derivative and that's what i got, idk if it's right tho

  10. anonymous
    • 5 years ago
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    So we can make it-------- -2x-4y-4xy'-6y^2y'=92. X=-2. Y=-4 So. 4+16+8y'-96y'=92 so y'=72/-88=-9/11

  11. anonymous
    • 5 years ago
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    Maybe I am wrong. It is too hard to write on iPad

  12. anonymous
    • 5 years ago
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    Sorry i am wrong. I forget to make 92 into 0

  13. anonymous
    • 5 years ago
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    yea so it would be -2x-4xy'=4y-6y^2=0

  14. anonymous
    • 5 years ago
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    sorry typo... i meant -2x-4xy'+4y-6y^2y'=0

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