Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
do we assume X = x ?
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
product rule it then and youll be good
 2 years ago

jrodriguez2315 Group TitleBest ResponseYou've already chosen the best response.1
Take the derivative of f'(x). Use product rule along with chain rule. f'(x)=f'(x)g(x) + f(x)g'(x).
 2 years ago

marmots Group TitleBest ResponseYou've already chosen the best response.0
I've tried that several times. I just need an answer.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
(3sqrtX)'(X^3 2sqrtX +6)+(3sqrtX)(X^3 2sqrtX +6)' \(\cfrac{3}{2\sqrt{x}}\) (X^3 2sqrtX +6)+ \(3\sqrt{x}\ (3x^2 \cfrac{1}{\sqrt{x}})\)
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{3x^3}{2\sqrt{x}}3+\frac{9}{\sqrt{x}}+9x^2\sqrt{x}3\]
 2 years ago

jrodriguez2315 Group TitleBest ResponseYou've already chosen the best response.1
marmots make sure to regard the sqrt(x) as (x)^1/2. It makes it easier to integrate. Just a tip.
 2 years ago

jrodriguez2315 Group TitleBest ResponseYou've already chosen the best response.1
I mean to take the derivative!!
 2 years ago

marmots Group TitleBest ResponseYou've already chosen the best response.0
Thanks guys, I appreciate it.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
yep, might need to simplify some more, but im pretty sure thats the brunt of it
 2 years 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.