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Study23 Group Title

PLEASE HELP! (Waiting for an hour..._) Find an equation of the tangent line to the curve at the given point?? \(\ \Large \frac{x^2}{16}-\frac{y^2}{9} =1\) \(\ \Large \text{The point is: } (-5, \frac{9}{4}) \). PLEASE HELP!

  • 2 years ago
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  1. jaersyn Group Title
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    what math are you in?

    • 2 years ago
  2. Study23 Group Title
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    AP Calculus BC

    • 2 years ago
  3. jaersyn Group Title
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    what's "BC"

    • 2 years ago
  4. Study23 Group Title
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    It's just a level of calculus.... I need to find the derivative of this equation...

    • 2 years ago
  5. jaersyn Group Title
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    solve for y and take the derivative?

    • 2 years ago
  6. Study23 Group Title
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    This section involves implicit differentiation

    • 2 years ago
  7. jaersyn Group Title
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    ohhh

    • 2 years ago
  8. satellite73 Group Title
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    did you take the derivative?

    • 2 years ago
  9. Study23 Group Title
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    No I'm still stuck on this problem!!

    • 2 years ago
  10. Study23 Group Title
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    I don't know how to with those fractions

    • 2 years ago
  11. satellite73 Group Title
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    the derivative of \(\frac{x^2}{16}\) is \(\frac{x}{8}\)

    • 2 years ago
  12. Study23 Group Title
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    Really? Wouldn't the 16 become a zero?

    • 2 years ago
  13. satellite73 Group Title
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    and the derivative with respect to \(x\) of \(-\frac{y^2}{9}\) is \[-\frac{2y}{9}y'\]

    • 2 years ago
  14. satellite73 Group Title
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    no it is a constant, think \[\frac{x^2}{16}=\frac{1}{16}x^2\]

    • 2 years ago
  15. Study23 Group Title
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    Am I just using power rule here? I don't need to use quotient rule (that's what I was thinking...?)

    • 2 years ago
  16. Study23 Group Title
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    @satellite73 I've been getting mixed answers from people. When do I know when I take the derivative of y to have yy' versus just y'???

    • 2 years ago
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