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

Calculus1
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Are you having issues finding y'?
Do you know the chain rule?
Ma'am will explain you, don't worry..

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Other answers:

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

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