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Dodo1 Group Title

d/dt ( 2^(t)2sin^(-1)(2t+1) derivatives.

  • one year ago
  • one year ago

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  1. Dodo1 Group Title
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    @tkhunny please help!

    • one year ago
  2. tkhunny Group Title
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    Looks like a good "paying attention" problem. What have you tried? This? \(\dfrac{d}{dt}2^{t}\cdot 2\cdot\sin^{-1}(2t+1)\) Personally, I would rewrite it just a little. This? \(\dfrac{d}{dt}2^{t+1}\cdot\sin^{-1}(2t+1)\)

    • one year ago
  3. Dodo1 Group Title
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    \[d/dt (2^t \sin ^{-1}(2t+1)\]

    • one year ago
  4. Dodo1 Group Title
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    product rule and chain rule?

    • one year ago
  5. tkhunny Group Title
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    Okay, so no extra 2 and there is no need or temptation to rewrite it. Great. You have the idea. Let's see what you get. Hint: \(\dfrac{d}{dt}2^{t} = 2^{t}\cdot\log(2)\)

    • one year ago
  6. Dodo1 Group Title
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    mmm i have no clue.

    • one year ago
  7. tkhunny Group Title
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    False. You had a clue a minute ago. Get it back.

    • one year ago
  8. Dodo1 Group Title
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    ok,

    • one year ago
  9. Dodo1 Group Title
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    chain rule for each numbers?

    • one year ago
  10. tkhunny Group Title
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    Hint: \(\dfrac{d}{dt}\sin^{-1}(t) = \dfrac{1}{\sqrt{1-t^{2}}}\) That's about all you need.

    • one year ago
  11. tkhunny Group Title
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    Stop asking questions and start writing the derivative.

    • one year ago
  12. tkhunny Group Title
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    \(\dfrac{d}{dt}f(t)\cdot g(t) = f(t)\cdot g'(t) + g(t)\cdot f'(t)\) Go!

    • one year ago
  13. Dodo1 Group Title
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    \[d/dt (2t+1)= (2)\]

    • one year ago
  14. tkhunny Group Title
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    That's part of it. Keep going.

    • one year ago
  15. Dodo1 Group Title
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    product rules? but it does not make sense \[2^t \sin^{-1} (2t+1) \] three parts. the product has two/ f(x)g(x)

    • one year ago
  16. Dodo1 Group Title
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    @tkhunny

    • one year ago
  17. tkhunny Group Title
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    If there where three, you could just group them and make pairs. There are only two. \(2^{t}\) \(\sin^{-1}(2t+1) = asin(2t+1)\)

    • one year ago
  18. Dodo1 Group Title
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    oh.. i see.

    • one year ago
  19. Dodo1 Group Title
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    \[2^t*(asin(2t+1)'+(asin(2t+1)(2^t)'\] ?

    • one year ago
  20. tkhunny Group Title
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    There's that question again. Move forward. Do NOT stop to ask every time you flinch! Keep going!

    • one year ago
  21. Dodo1 Group Title
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    I want to make sure that i am doing right. so i dont need to rise off all my effort and time writting wrong equation

    • one year ago
  22. tkhunny Group Title
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    Right. I understand your motivation. I am also telling you that it is holding you back. You have far too much fear. Learn the right things. Go the right direction. Go with confidence. What's next?

    • one year ago
  23. Dodo1 Group Title
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    \[(2^t)(1/\sqrt(1-t)*dx (2t+1)+asin(2t+1)(2^tlong(2)\]

    • one year ago
  24. tkhunny Group Title
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    So Close!! \(\dfrac{d}{dt}asin(t) = \dfrac{1}{\sqrt{1-t^{2}}}\) \(\dfrac{d}{dt}asin(2t+1) = \dfrac{1}{\sqrt{1-(2t+1)^{2}}}\cdot 2\)

    • one year ago
  25. tkhunny Group Title
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    See! That was very good. No need to fear. Just pay a little better attention.

    • one year ago
  26. Dodo1 Group Title
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    \[2^t \frac{ 2 }{ \sqrt{1-(2t+1)}}+(\sin^{-1}(2t+1)(2^t* \log (2)) \]

    • one year ago
  27. tkhunny Group Title
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    Missed the "squared" in the square root. Otherwise, perfect.

    • one year ago
  28. Dodo1 Group Title
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    sqrt(2t+1)?

    • one year ago
  29. tkhunny Group Title
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    Already showed you that. Pay better attention. \(\dfrac{2}{\sqrt{1 - (2t+1)^{2}}}\)

    • one year ago
  30. Dodo1 Group Title
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    ok, thank you

    • one year ago
  31. tkhunny Group Title
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    Now, seriously. This is where I look you in the eye and ask you if you are in the right class, if you are truly dedicated to this course, if you think you can gain the confidence to move forward - and various other questions about your mental health. Work hard. You'll get it!

    • one year ago
  32. Dodo1 Group Title
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    Thank you very much tkhunny. i will work hard and thank you for your advice.

    • one year ago
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