IsTim
  • IsTim
Use the first principles definition to determine the 1st derivative of h(t)=-2t/(t+3)
Mathematics
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SOLVED
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katieb
  • katieb
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da_ScienceMan
  • da_ScienceMan
d solution will not fir in here. add d appropriate delta change and simplify until u take limits. google might help lol
IsTim
  • IsTim
Sorry. I don't understand?
IsTim
  • IsTim
This is my furthest step: |dw:1340171061632:dw|

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da_ScienceMan
  • da_ScienceMan
m nt sure why u got so many square root signs u r meant to add small deltas to y and independent variable t
da_ScienceMan
  • da_ScienceMan
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
IsTim
  • IsTim
Sorry, I'm to use first principles here, not actual derivatives...
anonymous
  • anonymous
Here let me see if this helps answer your question... I got to type this up so give me a moment
da_ScienceMan
  • da_ScienceMan
i wrote it down but cannot sketch it
IsTim
  • IsTim
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.
IsTim
  • IsTim
USe the equation editor. That makes it look nicer.
da_ScienceMan
  • da_ScienceMan
\[(y + \delta y) = -2(t + \delta t) / (t + \delta t ) + 3\]
da_ScienceMan
  • da_ScienceMan
then next i say \[\delta y = -2(t + \delta t ) /(t + \delta t) + 3 - (2t / t + 3)\]
IsTim
  • IsTim
Just realized I was answering for the wrong question...
da_ScienceMan
  • da_ScienceMan
then u simplify further, until u nw say \[\delta y _{\lim_{\delta t \rightarrow 0}} = \]
da_ScienceMan
  • da_ScienceMan
the final answer u get on d RHS!
da_ScienceMan
  • da_ScienceMan
then u nw deal with the limits of course
anonymous
  • anonymous
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: \[\frac{-6}{(t+3)^2}\]
anonymous
  • anonymous
Are you needing this in the limit definition form?
IsTim
  • IsTim
Not sure.
da_ScienceMan
  • da_ScienceMan
first principles deals with limits
da_ScienceMan
  • da_ScienceMan
first principles translates to explicit use of differentiation from its basis as defined by first scientists lol
anonymous
  • anonymous
Because this gets VERY messy: \[\huge \lim_{{\triangle x} \rightarrow 0} \frac{f(x+\triangle x)+f(x)}{\triangle x}\]
IsTim
  • IsTim
I agree wholeheartedly.
anonymous
  • anonymous
Imagine taking the original function and putting (x+\(\triangle\)x) into each... eww
anonymous
  • anonymous
oh yeah and then adding to another function and dividing and looking for terms to cancel out
da_ScienceMan
  • da_ScienceMan
sure u have to do it patiently, until it gets to where u need to apply limits
IsTim
  • IsTim
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.
da_ScienceMan
  • da_ScienceMan
Sure
anonymous
  • anonymous
You're welcome! :-3

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