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

Mathematics
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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
Sorry. I don't understand?
This is my furthest step: |dw:1340171061632:dw|

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Other answers:

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

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