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Dodo1

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

  • 11 months ago
  • 11 months ago

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  1. Dodo1
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    @tkhunny please help!

    • 11 months ago
  2. tkhunny
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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? \(\dfrac{d}{dt}2^{t+1}\cdot\sin^{-1}(2t+1)\)

    • 11 months ago
  3. Dodo1
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    \[d/dt (2^t \sin ^{-1}(2t+1)\]

    • 11 months ago
  4. Dodo1
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    product rule and chain rule?

    • 11 months ago
  5. tkhunny
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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: \(\dfrac{d}{dt}2^{t} = 2^{t}\cdot\log(2)\)

    • 11 months ago
  6. Dodo1
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    mmm i have no clue.

    • 11 months ago
  7. tkhunny
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    False. You had a clue a minute ago. Get it back.

    • 11 months ago
  8. Dodo1
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    ok,

    • 11 months ago
  9. Dodo1
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    chain rule for each numbers?

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

    • 11 months ago
  11. tkhunny
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    Stop asking questions and start writing the derivative.

    • 11 months ago
  12. tkhunny
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    \(\dfrac{d}{dt}f(t)\cdot g(t) = f(t)\cdot g'(t) + g(t)\cdot f'(t)\) Go!

    • 11 months ago
  13. Dodo1
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    \[d/dt (2t+1)= (2)\]

    • 11 months ago
  14. tkhunny
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    That's part of it. Keep going.

    • 11 months ago
  15. Dodo1
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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)

    • 11 months ago
  16. Dodo1
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    @tkhunny

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

    • 11 months ago
  18. Dodo1
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    oh.. i see.

    • 11 months ago
  19. Dodo1
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    \[2^t*(asin(2t+1)'+(asin(2t+1)(2^t)'\] ?

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

    • 11 months ago
  21. Dodo1
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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

    • 11 months ago
  22. tkhunny
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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?

    • 11 months ago
  23. Dodo1
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    \[(2^t)(1/\sqrt(1-t)*dx (2t+1)+asin(2t+1)(2^tlong(2)\]

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

    • 11 months ago
  25. tkhunny
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    See! That was very good. No need to fear. Just pay a little better attention.

    • 11 months ago
  26. Dodo1
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    \[2^t \frac{ 2 }{ \sqrt{1-(2t+1)}}+(\sin^{-1}(2t+1)(2^t* \log (2)) \]

    • 11 months ago
  27. tkhunny
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    Missed the "squared" in the square root. Otherwise, perfect.

    • 11 months ago
  28. Dodo1
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    sqrt(2t+1)?

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

    • 11 months ago
  30. Dodo1
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    ok, thank you

    • 11 months ago
  31. tkhunny
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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!

    • 11 months ago
  32. Dodo1
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    Thank you very much tkhunny. i will work hard and thank you for your advice.

    • 11 months ago
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