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amy0799

  • 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. amistre64
    • one year ago
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    the slope of a line is its derivative ...

  2. amistre64
    • one year ago
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    d/dx (tangent line to a curve) = y', or slope of the curve

  3. amistre64
    • one year ago
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    right?

  4. amy0799
    • one year ago
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    yes, so I just need to take the derivative of x^2+xy-y^2 = 1?

  5. amistre64
    • one year ago
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    as i see it, yes

  6. amy0799
    • one year ago
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    i asked my other friends, and they took the 2nd derivative. so I'm now confused...

  7. amistre64
    • one year ago
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    \[y_{tan}=f'(a)(x-a)+f(b)\] \[\frac d{dx}y_{tan}=f'(a)\]

  8. amistre64
    • one year ago
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    the derivative of a line, is its slope

  9. amistre64
    • one year ago
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    |dw:1444585385770:dw|

  10. amistre64
    • one year ago
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    regardless of the curve ... the function we are taking the derivative of is a line

  11. amy0799
    • one year ago
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    \[m'(x)=\frac{ -2x-y }{ x-2y }\] is this correct?

  12. amistre64
    • one year ago
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    maybe .. now you gonna make me work lol x^2+xy-y^2 = 1 2xx'+x'y+xy' -2yy' = 0 2 3 2 3 4+3+2y' -6y' = 0 (2-6)y' = -7

  13. amistre64
    • one year ago
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    i tend to have less error when i plug in and then solve for y'

  14. amy0799
    • one year ago
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    i got 1.75

  15. amistre64
    • one year ago
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    7/4 yes

  16. amy0799
    • one year ago
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    thank you!

  17. amistre64
    • one year ago
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    good luck

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spraguer (Moderator)
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