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onegirl Group Title

Find an equation of the tangent line to y = f(x) at x = a. f(x) = sqrt(x^2 + 16), x = 13

  • one year ago
  • one year ago

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  1. SithsAndGiggles Group Title
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    \[f(x)=\sqrt{x^2+16}\Rightarrow f(13)=\sqrt{185}\\ f'(x)=\frac{2x}{\sqrt{x^2+16}}\Rightarrow f'(13)=\frac{26}{\sqrt{185}}\] The tangent line is given by the point-slope formula using the point (13, √185) and the slope 26/√185: \[y-\sqrt{185}=\frac{26}{\sqrt{185}}(x-13)\]

    • one year ago
  2. onegirl Group Title
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    so thats the equation? @SithAndGiggles

    • one year ago
  3. onegirl Group Title
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    @SithsAndGiggles

    • one year ago
  4. onegirl Group Title
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    @myko

    • one year ago
  5. myko Group Title
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    the derivative is wrong. The nominator should be just x

    • one year ago
  6. onegirl Group Title
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    okay u mean in 2x?

    • one year ago
  7. myko Group Title
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    yes

    • one year ago
  8. onegirl Group Title
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    ok

    • one year ago
  9. myko Group Title
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    x/sqrt(x^2+16)

    • one year ago
  10. onegirl Group Title
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    okay

    • one year ago
  11. onegirl Group Title
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    so it wouldn't be 26

    • one year ago
  12. myko Group Title
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    rest looks right, just change 26 for 13 and that's it

    • one year ago
  13. onegirl Group Title
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    okay thx

    • one year ago
  14. myko Group Title
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    yw

    • one year ago
  15. SithsAndGiggles Group Title
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    Thanks for catching the mistake

    • one year ago
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