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IsTim Group Title

Use the first principles definition to determine the 1st derivative of h(t)=-2t/(t+3)

  • 2 years ago
  • 2 years ago

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  1. da_ScienceMan Group Title
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    d solution will not fir in here. add d appropriate delta change and simplify until u take limits. google might help lol

    • 2 years ago
  2. IsTim Group Title
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    Sorry. I don't understand?

    • 2 years ago
  3. IsTim Group Title
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    This is my furthest step: |dw:1340171061632:dw|

    • 2 years ago
  4. da_ScienceMan Group Title
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    m nt sure why u got so many square root signs u r meant to add small deltas to y and independent variable t

    • 2 years ago
  5. da_ScienceMan Group Title
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    that is y + dy = -2(t + dt)/(t + dt ) + 3. Then do dy the subject and solve until u cannot simplify further. Then apply limits on both sides

    • 2 years ago
  6. IsTim Group Title
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    Sorry, I'm to use first principles here, not actual derivatives...

    • 2 years ago
  7. agentx5 Group Title
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    Here let me see if this helps answer your question... I got to type this up so give me a moment

    • 2 years ago
  8. da_ScienceMan Group Title
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    i wrote it down but cannot sketch it

    • 2 years ago
  9. IsTim Group Title
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    I understand how quotient and chain rule works at its simplest, but I am attempting to learn how to use first principles too. Sorry, I hadn't realized what you were doing until now.

    • 2 years ago
  10. IsTim Group Title
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    USe the equation editor. That makes it look nicer.

    • 2 years ago
  11. da_ScienceMan Group Title
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    \[(y + \delta y) = -2(t + \delta t) / (t + \delta t ) + 3\]

    • 2 years ago
  12. da_ScienceMan Group Title
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    then next i say \[\delta y = -2(t + \delta t ) /(t + \delta t) + 3 - (2t / t + 3)\]

    • 2 years ago
  13. IsTim Group Title
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    Just realized I was answering for the wrong question...

    • 2 years ago
  14. da_ScienceMan Group Title
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    then u simplify further, until u nw say \[\delta y _{\lim_{\delta t \rightarrow 0}} = \]

    • 2 years ago
  15. da_ScienceMan Group Title
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    the final answer u get on d RHS!

    • 2 years ago
  16. da_ScienceMan Group Title
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    then u nw deal with the limits of course

    • 2 years ago
  17. agentx5 Group Title
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    Step-by-step from here (check to see if this is the correct original intended problem) \[v(t) = h'(t) = \frac{d}{dt}\frac{-2t}{t+3} \] Factor out constants: \[-2 \frac{d}{dt}\frac{t}{t+3}\] Use the quotient rule, \[\frac{d}{dt}\frac{high}{low} = \frac{((low)d(high)-(high)d(low)}{(low)^2}\]where "high" = t and "low" = t+3: \[-2 \frac{(t+3) \frac{d}{dt}(t)-(t)\frac{d}{dt}(t+3)}{(t+3)^2}\] Differentiate the sum term by term (derivative of t is 1): \[-\frac{2 (t (-(\frac{d}{dt}(t)+\frac{d}{dt}(3)))+t+3)}{(t+3)^2}\] The derivative of 3 is zero: \[-\frac{2 (t (-(d/dt(t)+0))+t+3)}{(t+3)^2}\] The derivative of t is 1: \[\frac{-6}{(t+3)^2}\]

    • 2 years ago
  18. agentx5 Group Title
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    Are you needing this in the limit definition form?

    • 2 years ago
  19. IsTim Group Title
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    Not sure.

    • 2 years ago
  20. da_ScienceMan Group Title
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    first principles deals with limits

    • 2 years ago
  21. da_ScienceMan Group Title
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    first principles translates to explicit use of differentiation from its basis as defined by first scientists lol

    • 2 years ago
  22. agentx5 Group Title
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    Because this gets VERY messy: \[\huge \lim_{{\triangle x} \rightarrow 0} \frac{f(x+\triangle x)+f(x)}{\triangle x}\]

    • 2 years ago
  23. IsTim Group Title
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    I agree wholeheartedly.

    • 2 years ago
  24. agentx5 Group Title
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    Imagine taking the original function and putting (x+\(\triangle\)x) into each... eww

    • 2 years ago
  25. agentx5 Group Title
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    oh yeah and then adding to another function and dividing and looking for terms to cancel out

    • 2 years ago
  26. da_ScienceMan Group Title
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    sure u have to do it patiently, until it gets to where u need to apply limits

    • 2 years ago
  27. IsTim Group Title
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    Thank you all for responding. The information, even if it didn't pertain to the question exactly, will help me further in my review. Yet again, thank you.

    • 2 years ago
  28. da_ScienceMan Group Title
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    Sure

    • 2 years ago
  29. agentx5 Group Title
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    You're welcome! :-3

    • 2 years ago
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