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monroe17

  • 2 years ago

Find an equation of the tangent line to the curve at the point. x^2+xy+y^2=7 (2,1) please show steps! I need to learn how to do it

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  1. ChmE
    • 2 years ago
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    Have you been taught derivatives

  2. monroe17
    • 2 years ago
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    yes, but tomorrow I have an exam on a bunch of stuf and im sooooooo lost!!! I'm just trying to learn it all because I don't understand my professor at all.

  3. ChmE
    • 2 years ago
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    okay lets start with this problem. Every time you take the derivative of a y we need to add y'. Ex) the derivative of 2y^2=6 is 4yy'=0, then you solve for y' We will do the same above. Let's work one term at a time. What is the derivative of x^2?

  4. monroe17
    • 2 years ago
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    2x

  5. timo86m
    • 2 years ago
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    Do implicit differentiation in this one i suppose. do it for both y and x

  6. monroe17
    • 2 years ago
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    thats the prob idk how

  7. timo86m
    • 2 years ago
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    \[-{\frac {x+2\,y}{2\,x+y}}\]

  8. timo86m
    • 2 years ago
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    http://www.khanacademy.org/math/calculus/differential-calculus/v/implicit-differentiation this could help

  9. timo86m
    • 2 years ago
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    http://www.khanacademy.org/math/calculus/differential-calculus/v/implicit-differentiation--part-2 part 2 is nice too

  10. timo86m
    • 2 years ago
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    well when y is dependant variable -(2*x+y)/(x+2*y)

  11. timo86m
    • 2 years ago
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    at 2,1 the slope is -5 -- 4

  12. monroe17
    • 2 years ago
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    I figure it out. But, now I'm stuck on the next problem

  13. timo86m
    • 2 years ago
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    use y=mx+b 1=-5/4*2+b sfb

  14. timo86m
    • 2 years ago
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    b=7/2 so y=-5/4*x+ 7/2 :)

  15. timo86m
    • 2 years ago
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    BTW I graphed it and it works fine both the tangent line and the function yyou gave me do you wanna see it?

  16. monroe17
    • 2 years ago
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    yeah i already figured it out

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