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

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- El_Tucan

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

- schrodinger

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- myininaya

Are you having issues finding y'?

- myininaya

Do you know the chain rule?

- anonymous

Ma'am will explain you, don't worry..

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## More answers

- El_Tucan

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

y'=f(x)
y'=0 implies you set f(x)=0 and solve for x

- myininaya

Just set the derivative equal to zero and solve for x

- myininaya

on the interval [0,pi)

- El_Tucan

i have y' = 2x * cos(x^2)
correct?
so 2x * cos(x^2) = 0?

- myininaya

What did you get for y'?

- myininaya

Right! :)

- myininaya

So we have x=0 or cos(x^2)=0

- myininaya

You need to solve cos(x^2)=0 for x on [0,pi)

- El_Tucan

i plugged cos(0^2) in my graphing calculator which is 1, is that right? im sorry :(

- anonymous

You are to find values for which it becomes 0 not 1..

- myininaya

For what values satisfy cos(u)=0
look at unit circle.

- El_Tucan

gotcha, thanks :)

- anonymous

Really??

- myininaya

You are actually solving cos(u)=0 for u on 0

- El_Tucan

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

- anonymous

Sorry for what dear??
I was just confirming that...

- El_Tucan

i thought you were picking on me :)
im working on my math esteem lol

- anonymous

There is nothing like that, I was just confirming that you got it or not...

- myininaya

Hey @El_Tucan that 8pi/2 should be something else

- myininaya

Those other two values you got were right with the little square root thingy that you said :)

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