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Dodo1

  • 2 years ago

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

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  1. Dodo1
    • 2 years ago
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    @tkhunny please help!

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

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

  4. Dodo1
    • 2 years ago
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    product rule and chain rule?

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

  6. Dodo1
    • 2 years ago
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    mmm i have no clue.

  7. tkhunny
    • 2 years ago
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    False. You had a clue a minute ago. Get it back.

  8. Dodo1
    • 2 years ago
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    ok,

  9. Dodo1
    • 2 years ago
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    chain rule for each numbers?

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

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

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

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

  14. tkhunny
    • 2 years ago
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    That's part of it. Keep going.

  15. Dodo1
    • 2 years ago
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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)

  16. Dodo1
    • 2 years ago
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    @tkhunny

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

  18. Dodo1
    • 2 years ago
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    oh.. i see.

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

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

  21. Dodo1
    • 2 years ago
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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

  22. tkhunny
    • 2 years ago
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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?

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

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

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

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

  27. tkhunny
    • 2 years ago
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    Missed the "squared" in the square root. Otherwise, perfect.

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

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

  30. Dodo1
    • 2 years ago
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    ok, thank you

  31. tkhunny
    • 2 years ago
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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!

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

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