Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
Can someone help me finish this derivative, using the first principle rule. [picture attached]
 10 months ago
 10 months ago
Can someone help me finish this derivative, using the first principle rule. [picture attached]
 10 months ago
 10 months ago

This Question is Closed

burhan101Best ResponseYou've already chosen the best response.0
dw:1371172930181:dw
 10 months ago

CallistoBest ResponseYou've already chosen the best response.0
\[f(t) = \frac{2t}{t+3}\]\[f'(t) = \lim_{h \rightarrow 0}\frac{f(t+h)f(t)}{h}\] \[f(t+h) =\frac{2(t+h)}{(t+h)+3}\] Does that help?
 10 months ago

burhan101Best ResponseYou've already chosen the best response.0
is what i have done so far right ? (there's an attachment)
 10 months ago

whpalmer4Best ResponseYou've already chosen the best response.1
\[\lim_{h>0}\frac{1}{h}*[\frac{6h6t2ht2t^2}{(t+3)(t+h+3)}+\frac{6t+2ht+2t^2}{(t+3)(t+h+3)}]\] is what you get after a bit of the algebraic dust settles...
 10 months ago

burhan101Best ResponseYou've already chosen the best response.0
that's exactly what im getting but i'm not understanding why my answer key shows \[\frac{ 5t2 }{ 2(5t1)^{\frac{ 3 }{ 2 }} }\]
 10 months ago

CallistoBest ResponseYou've already chosen the best response.0
Double check the answer key?
 10 months ago

burhan101Best ResponseYou've already chosen the best response.0
I have, maybe it's just wrong :S
 10 months ago

whpalmer4Best ResponseYou've already chosen the best response.1
That's the right answer, if the problem is to find the derivative of \[\frac{t}{\sqrt{5t1}}\]
 10 months ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.