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

1. myininaya

Are you having issues finding y'?

2. myininaya

Do you know the chain rule?

3. waterineyes

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

4. 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"

5. myininaya

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

6. myininaya

Just set the derivative equal to zero and solve for x

7. myininaya

on the interval [0,pi)

8. El_Tucan

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

9. myininaya

What did you get for y'?

10. myininaya

Right! :)

11. myininaya

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

12. myininaya

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

13. El_Tucan

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

14. waterineyes

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

15. myininaya

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

16. El_Tucan

gotcha, thanks :)

17. waterineyes

Really??

18. myininaya

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

19. 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

20. waterineyes

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

21. El_Tucan

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

22. waterineyes

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

23. myininaya

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

24. myininaya

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