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

1. da_ScienceMan

d solution will not fir in here. add d appropriate delta change and simplify until u take limits. google might help lol

2. IsTim

Sorry. I don't understand?

3. IsTim

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

4. 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

5. 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

6. IsTim

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

7. anonymous

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

8. da_ScienceMan

i wrote it down but cannot sketch it

9. 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.

10. IsTim

USe the equation editor. That makes it look nicer.

11. da_ScienceMan

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

12. da_ScienceMan

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

13. IsTim

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

14. da_ScienceMan

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

15. da_ScienceMan

the final answer u get on d RHS!

16. da_ScienceMan

then u nw deal with the limits of course

17. 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: $\frac{-6}{(t+3)^2}$

18. anonymous

Are you needing this in the limit definition form?

19. IsTim

Not sure.

20. da_ScienceMan

first principles deals with limits

21. da_ScienceMan

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

22. anonymous

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

23. IsTim

I agree wholeheartedly.

24. anonymous

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

25. anonymous

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

26. da_ScienceMan

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

27. 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.

28. da_ScienceMan

Sure

29. anonymous

You're welcome! :-3