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El_Tucan
 2 years ago
find all values in [0,pi) so that dy/dx = 0 for the function y = sin(x^2)
El_Tucan
 2 years ago
find all values in [0,pi) so that dy/dx = 0 for the function y = sin(x^2)

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

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

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

El_Tucan
 2 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
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.1Just set the derivative equal to zero and solve for x

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

El_Tucan
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.1What did you get for y'?

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

myininaya
 2 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
 2 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 :(

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

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

myininaya
 2 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
 2 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

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

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

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

myininaya
 2 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 :)
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