El_Tucan
find all values in [0,pi) so that dy/dx = 0 for the function y = sin(x^2)
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myininaya
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Are you having issues finding y'?
myininaya
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Do you know the chain rule?
waterineyes
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Ma'am will explain you, don't worry..
El_Tucan
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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"
myininaya
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y'=f(x)
y'=0 implies you set f(x)=0 and solve for x
myininaya
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Just set the derivative equal to zero and solve for x
myininaya
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on the interval [0,pi)
El_Tucan
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i have y' = 2x * cos(x^2)
correct?
so 2x * cos(x^2) = 0?
myininaya
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What did you get for y'?
myininaya
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Right! :)
myininaya
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So we have x=0 or cos(x^2)=0
myininaya
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You need to solve cos(x^2)=0 for x on [0,pi)
El_Tucan
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i plugged cos(0^2) in my graphing calculator which is 1, is that right? im sorry :(
waterineyes
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You are to find values for which it becomes 0 not 1..
myininaya
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For what values satisfy cos(u)=0
look at unit circle.
El_Tucan
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gotcha, thanks :)
waterineyes
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Really??
myininaya
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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
El_Tucan
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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
waterineyes
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Sorry for what dear??
I was just confirming that...
El_Tucan
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i thought you were picking on me :)
im working on my math esteem lol
waterineyes
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There is nothing like that, I was just confirming that you got it or not...
myininaya
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Hey @El_Tucan that 8pi/2 should be something else
myininaya
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Those other two values you got were right with the little square root thingy that you said :)