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anonymous

  • one year ago

d/dx m(x) where m(x) is the tangent line to curve x^2+xy-y^2=1 at point (2,3)

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  1. anonymous
    • one year ago
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    my work: \( 2x+y+xy'-2yy'=0\\ 2x+y=2yy'-xy'\\ 2x+y=y'(2y-x)\\ \dfrac{dy}{dx}=\dfrac{2x+y}{2y-x} \)

  2. anonymous
    • one year ago
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    do i just plug the point in now? or do I use y-y1=m(x-x1) somehow

  3. mathmate
    • one year ago
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    Since m'(x)=the slope m at (2,3), y'(2) should equal m'(x).

  4. mathmate
    • one year ago
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    * y'(2,3)

  5. anonymous
    • one year ago
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    I'm not sure I understand what the question is asking? is 7 the answer?

  6. Loser66
    • one year ago
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    \(\dfrac{d}{dx}(m(x))\) means : after you got m(x), take derivative of it again.

  7. anonymous
    • one year ago
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    ohhh

  8. Loser66
    • one year ago
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    and then plug (2,3) in

  9. anonymous
    • one year ago
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    it's been like 2 years since I took a calc class, was helping someone with this question. thank you both

  10. anonymous
    • one year ago
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    thank you https://www.desmos.com/calculator/ykoruebu4u

  11. mathmate
    • one year ago
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    |dw:1444436902647:dw| The question is asking for the slope of the tangent line m(x), which should be the same as the slope of the curve at (2,3), by definition of the tangent. that is why m'(x)=slope of tangent line = y'(2,3)

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