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youngbuck2123 Group Title

How would you use implicit differentiation to find an equation of the tangent line to the curve x^2+ 2xy-y^2+x=2 at the point (1,2)

  • 2 years ago
  • 2 years ago

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  1. myininaya Group Title
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    First step: Take derivative with respect to x on both sides of the equation.

    • 2 years ago
  2. myininaya Group Title
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    \[ (x^2+2xy-y^2+x)'=(2)'\] ----- \[(x^2+2xy-y^2+x)'=(x^2)'+(2xy)'-(y^2)'+(x)'=....\] \[\text{Here you will need power rule for } (x^2)'\] \[\text{You will need constant multiple rule along with product rule for } (2xy)'\] \[\text{You will need chain rule for } (y^2)'\] \[\text{You will need to know that }(x)'=1 \text{ for the } (x)' \text{part}\] ---- \[(2)'=....\] You should know the derivative of a constant. :) ----

    • 2 years ago
  3. youngbuck2123 Group Title
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    i understand the steps i just cannot put 2 and 2 together and get the correct final answer. i got 1/3 as the slope, is the correct?

    • 2 years ago
  4. myininaya Group Title
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    Hmm...Not getting 1/3.... Show me your work so I can see where you messed up. You can do it one line at a time if you want.

    • 2 years ago
  5. myininaya Group Title
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    \[\text{What did you get when you did } (2xy)' \text{ and } (y^2)' ?\]

    • 2 years ago
  6. youngbuck2123 Group Title
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    (2xy)' = dy/dx 2y +2x dy/dx and (y^2)' = -2y

    • 2 years ago
  7. myininaya Group Title
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    Looks incorrect for both

    • 2 years ago
  8. myininaya Group Title
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    \[(2xy)' =2(xy)' \text{ By constant multiple rule}\] \[2(xy)'=2[xy]'=2[(x)'y+x(y)'] \text{ By product rule }\]

    • 2 years ago
  9. myininaya Group Title
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    \[(y^n)'=n y ^{n-1} y' \text{ By chain rule } \]

    • 2 years ago
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