Question

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

Answer 1

Are you having issues finding y'?

Answer 2

Do you know the chain rule?

Answer 3

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

Answer 4

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"

Answer 5

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

Answer 6

Just set the derivative equal to zero and solve for x

Answer 7

on the interval [0,pi)

Answer 8

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

Answer 9

What did you get for y'?

Answer 10

Right! :)

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

gotcha, thanks :)

Answer 17

Really??

Answer 18

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

Answer 19

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

Answer 20

Sorry for what dear??

I was just confirming that...

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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