- IsTim

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

- jamiebookeater

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- 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

Sorry. I don't understand?

- IsTim

This is my furthest step:
|dw:1340171061632:dw|

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## More answers

- 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

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

Sorry, I'm to use first principles here, not actual derivatives...

- anonymous

Here let me see if this helps answer your question... I got to type this up so give me a moment

- da_ScienceMan

i wrote it down but cannot sketch it

- 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

USe the equation editor. That makes it look nicer.

- da_ScienceMan

\[(y + \delta y) = -2(t + \delta t) / (t + \delta t ) + 3\]

- da_ScienceMan

then next i say \[\delta y = -2(t + \delta t ) /(t + \delta t) + 3 - (2t / t + 3)\]

- IsTim

Just realized I was answering for the wrong question...

- da_ScienceMan

then u simplify further, until u nw say \[\delta y _{\lim_{\delta t \rightarrow 0}} = \]

- da_ScienceMan

the final answer u get on d RHS!

- da_ScienceMan

then u nw deal with the limits of course

- 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

Are you needing this in the limit definition form?

- IsTim

Not sure.

- da_ScienceMan

first principles deals with limits

- da_ScienceMan

first principles translates to explicit use of differentiation from its basis as defined by first scientists lol

- anonymous

Because this gets VERY messy:
\[\huge \lim_{{\triangle x} \rightarrow 0} \frac{f(x+\triangle x)+f(x)}{\triangle x}\]

- IsTim

I agree wholeheartedly.

- anonymous

Imagine taking the original function and putting (x+\(\triangle\)x) into each... eww

- anonymous

oh yeah and then adding to another function and dividing and looking for terms to cancel out

- da_ScienceMan

sure u have to do it patiently, until it gets to where u need to apply limits

- 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

Sure

- anonymous

You're welcome! :-3

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