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Dodo1

  • one year ago

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

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  1. Dodo1
    • one year ago
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    @tkhunny please help!

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

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

  5. tkhunny
    • one year 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
    • one year ago
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    mmm i have no clue.

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

  8. Dodo1
    • one year ago
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    ok,

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

  10. tkhunny
    • one year 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
    • one year ago
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    Stop asking questions and start writing the derivative.

  12. tkhunny
    • one year 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
    • one year ago
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    \[d/dt (2t+1)= (2)\]

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

  15. Dodo1
    • one year 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
    • one year ago
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    @tkhunny

  17. tkhunny
    • one year 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
    • one year ago
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    oh.. i see.

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

  20. tkhunny
    • one year 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
    • one year 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
    • one year 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
    • one year ago
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    \[(2^t)(1/\sqrt(1-t)*dx (2t+1)+asin(2t+1)(2^tlong(2)\]

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

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

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

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

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

  30. Dodo1
    • one year ago
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    ok, thank you

  31. tkhunny
    • one year 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
    • one year ago
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    Thank you very much tkhunny. i will work hard and thank you for your advice.

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