A community for students.
Here's the question you clicked on:
 0 viewing
bigcat
 3 years ago
find the derivative of f(x)=x^2 + x 3 using the definition of the derivative
bigcat
 3 years ago
find the derivative of f(x)=x^2 + x 3 using the definition of the derivative

This Question is Closed

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.3\[f'(x)\]\[=\lim_{h \rightarrow 0}\frac{f(x+h)  f(x)}{h}\]\[=\lim_{h \rightarrow 0}\frac{[(x+h)^2 + (x+h) 3] (x^2 + x 3)}{h}\]Can you simplify the fraction first?

bigcat
 3 years ago
Best ResponseYou've already chosen the best response.0I used x^2 + 2(x)(h) + h^2 + x +h 3 x^2 x +3 = 2x + x ??

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.3x^2 + 2(x)(h) + h^2 + x +h 3 x^2 x +3 < correct for denominator. But it's not equal to 2x + x... \[x^2 + 2(x)(h) + h^2 + x +h 3 x^2 x +3\]Group the like terms together: \[x^2 x^2 + x x 3 +3+ h^2+ 2(x)(h)+h\] Now, can you simplify the above expression

bigcat
 3 years ago
Best ResponseYou've already chosen the best response.0basically factor out the h and H(h+2x+1) divided by the h on denominator and = h + 2x + 1?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.3And you got: \[\lim_{h \rightarrow 0} (h+2x+1)\]Can you evaluate the limit?

bigcat
 3 years ago
Best ResponseYou've already chosen the best response.0Would be 2x + 1 ??? Because h is going to 0

bigcat
 3 years ago
Best ResponseYou've already chosen the best response.0Holy Cow! It came together finally!! Thank you very much for your help! I would have been stuck!!!
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.