A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
the derivative of 2x^2
I need to know how to solve this using the derivative as a function... (using limits)
anonymous
 5 years ago
the derivative of 2x^2 I need to know how to solve this using the derivative as a function... (using limits)

This Question is Closed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1\[\lim_{h>0}\frac{f(x+h)f(x)}{h}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I already know that formula, i need to know how to use it....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1\[\lim\frac{2(x+h)^22x^2}{h}\] \[\lim\frac{2(x^2+2xh+h^2)2x^2}{h}\] \[\lim\frac{2x^2+4xh+2h^22x^2}{h}\] \[\lim\frac{\cancel{2x^2}+4xh+2h^2\cancel{2x^2}}{h}\] \[\lim\frac{4xh}{h}+\frac{2h^2}{h}\] \[\lim\frac{4x}{1}+\frac{2h}{1}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1if you know the formula, you know how to use it :/

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1now, when h=0; 4x+2h = 4x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1when I first did these the "x" part of the function confused me a bit since they used x as part of the substituions

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1f(x+h) should be written better as say: f(a+h) maybe so that you know to replace "x" by whats in the ( )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, now i get it thanks alot

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1youre welcome, and good luck :)
Ask your own question
Sign UpFind more explanations on OpenStudy
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.