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

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da_ScienceManBest 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
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
Sorry. I don't understand?
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
This is my furthest step: dw:1340171061632:dw
 one year ago

da_ScienceManBest 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
 one year ago

da_ScienceManBest 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
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
Sorry, I'm to use first principles here, not actual derivatives...
 one year ago

agentx5Best 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
 one year ago

da_ScienceManBest ResponseYou've already chosen the best response.1
i wrote it down but cannot sketch it
 one year ago

IsTimBest 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.
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
USe the equation editor. That makes it look nicer.
 one year ago

da_ScienceManBest ResponseYou've already chosen the best response.1
\[(y + \delta y) = 2(t + \delta t) / (t + \delta t ) + 3\]
 one year ago

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

IsTimBest ResponseYou've already chosen the best response.0
Just realized I was answering for the wrong question...
 one year ago

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

da_ScienceManBest ResponseYou've already chosen the best response.1
the final answer u get on d RHS!
 one year ago

da_ScienceManBest ResponseYou've already chosen the best response.1
then u nw deal with the limits of course
 one year ago

agentx5Best 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: \[\frac{6}{(t+3)^2}\]
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Are you needing this in the limit definition form?
 one year ago

da_ScienceManBest ResponseYou've already chosen the best response.1
first principles deals with limits
 one year ago

da_ScienceManBest 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
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Because this gets VERY messy: \[\huge \lim_{{\triangle x} \rightarrow 0} \frac{f(x+\triangle x)+f(x)}{\triangle x}\]
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Imagine taking the original function and putting (x+\(\triangle\)x) into each... eww
 one year ago

agentx5Best 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
 one year ago

da_ScienceManBest 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
 one year ago

IsTimBest 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.
 one year ago
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