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anonymous

  • 5 years ago

find the slope of the tangent line to the curve -1x^2-4xy-2y^3=92 at point (-2,-4)

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  1. anonymous
    • 5 years ago
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    Use some implicit differentiation, differentiate the function, solve for dy/dx and plug in the points (-2,-4). Then, just use y-y1=m(x-x1) to find the tangent line.

  2. anonymous
    • 5 years ago
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    yes i get the concept and i did all that but, it's saying i have the wrong answer :(

  3. anonymous
    • 5 years ago
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    Hm...can you use the equation feature to clean up the function for me? Then I'll try it out and tell you what I get. :)

  4. anonymous
    • 5 years ago
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    i got -2x-4x y'+4x' y-6y^2 y'=0

  5. anonymous
    • 5 years ago
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    Mainly the first term...is it -x^2?

  6. anonymous
    • 5 years ago
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    yes but they wrote it as -1x^2

  7. anonymous
    • 5 years ago
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    i got y'=2x-4y/-4x-6y^2

  8. anonymous
    • 5 years ago
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    sot sure if that's right

  9. anonymous
    • 5 years ago
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    I'm getting this: \[-2x - 4x*\frac{dy}{dx} - 4y - 6y^2\frac{dy}{dx} = 0\]\[-2x-4y = 4x \frac{dy}{dx} + 6y^2\frac{dy}{dx}\]\[\frac{dy}{dx} = \frac{-2x-4y}{4x+6y^2}.\]

  10. anonymous
    • 5 years ago
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    sorry i couldn't read what u got

  11. anonymous
    • 5 years ago
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    ...hm?

  12. anonymous
    • 5 years ago
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    sorry what u sent came up as a math processing error on my screen, but i g2g so thanks anyways

  13. anonymous
    • 5 years ago
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    maybe next time

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spraguer (Moderator)
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