El_Tucan
  • El_Tucan
find all values in [0,pi) so that dy/dx = 0 for the function y = sin(x^2)
Calculus1
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SOLVED
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schrodinger
  • schrodinger
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myininaya
  • myininaya
Are you having issues finding y'?
myininaya
  • myininaya
Do you know the chain rule?
anonymous
  • anonymous
Ma'am will explain you, don't worry..

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

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