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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
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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da_ScienceMan Group TitleBest ResponseYou've already chosen the best response.1
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

IsTim Group TitleBest ResponseYou've already chosen the best response.0
Sorry. I don't understand?
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
This is my furthest step: dw:1340171061632:dw
 2 years ago

da_ScienceMan Group TitleBest ResponseYou've already chosen the best response.1
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

da_ScienceMan Group TitleBest ResponseYou've already chosen the best response.1
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

IsTim Group TitleBest ResponseYou've already chosen the best response.0
Sorry, I'm to use first principles here, not actual derivatives...
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Here let me see if this helps answer your question... I got to type this up so give me a moment
 2 years ago

da_ScienceMan Group TitleBest ResponseYou've already chosen the best response.1
i wrote it down but cannot sketch it
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
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

IsTim Group TitleBest ResponseYou've already chosen the best response.0
USe the equation editor. That makes it look nicer.
 2 years ago

da_ScienceMan Group TitleBest ResponseYou've already chosen the best response.1
\[(y + \delta y) = 2(t + \delta t) / (t + \delta t ) + 3\]
 2 years ago

da_ScienceMan Group TitleBest ResponseYou've already chosen the best response.1
then next i say \[\delta y = 2(t + \delta t ) /(t + \delta t) + 3  (2t / t + 3)\]
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
Just realized I was answering for the wrong question...
 2 years ago

da_ScienceMan Group TitleBest ResponseYou've already chosen the best response.1
then u simplify further, until u nw say \[\delta y _{\lim_{\delta t \rightarrow 0}} = \]
 2 years ago

da_ScienceMan Group TitleBest ResponseYou've already chosen the best response.1
the final answer u get on d RHS!
 2 years ago

da_ScienceMan Group TitleBest ResponseYou've already chosen the best response.1
then u nw deal with the limits of course
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Stepbystep 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

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Are you needing this in the limit definition form?
 2 years ago

da_ScienceMan Group TitleBest ResponseYou've already chosen the best response.1
first principles deals with limits
 2 years ago

da_ScienceMan Group TitleBest ResponseYou've already chosen the best response.1
first principles translates to explicit use of differentiation from its basis as defined by first scientists lol
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Because this gets VERY messy: \[\huge \lim_{{\triangle x} \rightarrow 0} \frac{f(x+\triangle x)+f(x)}{\triangle x}\]
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
I agree wholeheartedly.
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Imagine taking the original function and putting (x+\(\triangle\)x) into each... eww
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
oh yeah and then adding to another function and dividing and looking for terms to cancel out
 2 years ago

da_ScienceMan Group TitleBest ResponseYou've already chosen the best response.1
sure u have to do it patiently, until it gets to where u need to apply limits
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
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

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
You're welcome! :3
 2 years ago
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