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

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

waterineyes
 one year ago
Best ResponseYou've already chosen the best response.0Ma'am will explain you, don't worry..

El_Tucan
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1Just set the derivative equal to zero and solve for x

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

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

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

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

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

El_Tucan
 one year 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 :(

waterineyes
 one year ago
Best ResponseYou've already chosen the best response.0You are to find values for which it becomes 0 not 1..

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

myininaya
 one year 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
 one year 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

waterineyes
 one year ago
Best ResponseYou've already chosen the best response.0Sorry for what dear?? I was just confirming that...

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

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

myininaya
 one year 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 :)
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