A community for students.
Here's the question you clicked on:
 0 viewing
El_Tucan
 3 years ago
find all values in [0,pi) so that dy/dx = 0 for the function y = sin(x^2)
El_Tucan
 3 years ago
find all values in [0,pi) so that dy/dx = 0 for the function y = sin(x^2)

This Question is Closed

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1Are you having issues finding y'?

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1Do you know the chain rule?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ma'am will explain you, don't worry..

El_Tucan
 3 years ago
Best ResponseYou've already chosen the best response.0yeah i get using the chain rule to find the derivative but i don't know what is meant by "finding values so that dy/dx = 0"

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1y'=f(x) y'=0 implies you set f(x)=0 and solve for x

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1Just set the derivative equal to zero and solve for x

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1on the interval [0,pi)

El_Tucan
 3 years ago
Best ResponseYou've already chosen the best response.0i have y' = 2x * cos(x^2) correct? so 2x * cos(x^2) = 0?

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1What did you get for y'?

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1So we have x=0 or cos(x^2)=0

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1You need to solve cos(x^2)=0 for x on [0,pi)

El_Tucan
 3 years ago
Best ResponseYou've already chosen the best response.0i plugged cos(0^2) in my graphing calculator which is 1, is that right? im sorry :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You are to find values for which it becomes 0 not 1..

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1For what values satisfy cos(u)=0 look at unit circle.

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1You are actually solving cos(u)=0 for u on 0<u<=pi^2 Keep in mind that pi^2 is between 3pi and 7 pi/2 So you actually want to find values for u that satisfy cos(u)=0 on 0<u<3pi

El_Tucan
 3 years ago
Best ResponseYou've already chosen the best response.0sorry waterineyes ;) i got it, thank you myininaya, the square roots of pi/2, 8pi/2, & 5pi/2 are the zero's within that interval

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry for what dear?? I was just confirming that...

El_Tucan
 3 years ago
Best ResponseYou've already chosen the best response.0i thought you were picking on me :) im working on my math esteem lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There is nothing like that, I was just confirming that you got it or not...

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1Hey @El_Tucan that 8pi/2 should be something else

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1Those other two values you got were right with the little square root thingy that you said :)
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.