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anonymous

  • 5 years ago

dx/dt of ln((t^2)+3)

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  1. myininaya
    • 5 years ago
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    \[(\ln(t^2+3))'=\frac{(t^2+3)'}{t^2+3}\]

  2. anonymous
    • 5 years ago
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    how about if we dash it again? what would the answer be?

  3. anonymous
    • 5 years ago
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    use product rule.

  4. anonymous
    • 5 years ago
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    eerrrr quotient rule i mean

  5. myininaya
    • 5 years ago
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    you mean chain rule

  6. anonymous
    • 5 years ago
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    could you show me how to do that please?

  7. myininaya
    • 5 years ago
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    i showed you above all you need to do is find the derivative of t^2 and derivative of 3 \[\frac{(t^2+3)'}{t^2+3}=\frac{2t+0}{t^2+3}\]

  8. anonymous
    • 5 years ago
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    oh no what i mean is if we had to dash it again , like second derivative

  9. myininaya
    • 5 years ago
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    dash it again? so you want to find \[(\frac{2t}{t^2+3})'?\]

  10. myininaya
    • 5 years ago
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    you will use quotient rule

  11. anonymous
    • 5 years ago
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    yes

  12. anonymous
    • 5 years ago
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    derivative ln(t2+3) 1/(t^2+3)*(2t)Dt

  13. anonymous
    • 5 years ago
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    dash again

  14. anonymous
    • 5 years ago
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    |dw:1326853065865:dw|

  15. anonymous
    • 5 years ago
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    i want to find second derivative

  16. myininaya
    • 5 years ago
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    shiign just put 2nd derivative

  17. myininaya
    • 5 years ago
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    all he did was use quotient rule

  18. anonymous
    • 5 years ago
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    so the answer is -2t^2 +6 /(t^2+3)^2

  19. anonymous
    • 5 years ago
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    is that correct or not?

  20. anonymous
    • 5 years ago
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    yes

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