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

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

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

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

El_TucanBest ResponseYou've already chosen the best response.0
yeah 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"
 one year ago

myininayaBest ResponseYou've already chosen the best response.1
y'=f(x) y'=0 implies you set f(x)=0 and solve for x
 one year ago

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

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

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

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

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

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

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

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

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

myininayaBest ResponseYou've already chosen the best response.1
You 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
 one year ago

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

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

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

waterineyesBest ResponseYou've already chosen the best response.0
There is nothing like that, I was just confirming that you got it or not...
 one year ago

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

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