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

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da_ScienceMan
 4 years ago
Best ResponseYou've already chosen the best response.1d solution will not fir in here. add d appropriate delta change and simplify until u take limits. google might help lol

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry. I don't understand?

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

da_ScienceMan
 4 years ago
Best ResponseYou've already chosen the best response.1m 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
 4 years ago
Best ResponseYou've already chosen the best response.1that 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
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, I'm to use first principles here, not actual derivatives...

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

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

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0USe the equation editor. That makes it look nicer.

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

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

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

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Stepbystep 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
 4 years ago
Best ResponseYou've already chosen the best response.0Are you needing this in the limit definition form?

da_ScienceMan
 4 years ago
Best ResponseYou've already chosen the best response.1first principles deals with limits

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Imagine taking the original function and putting (x+\(\triangle\)x) into each... eww

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh yeah and then adding to another function and dividing and looking for terms to cancel out

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

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0Thank 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.
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