y` is y prime

I don't think that is right...the dx/dy is what is throwing me off...

quote:

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•Delidgamond• said this about your mom:
take the derivitive of sin^2(x-3y)
--------------
2sin(x-3y) * (1-3y`) = dx/dy

y` is y prime

I don't think that is right...the dx/dy is what is throwing me off...

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It should equal 0.

So, you will get 2sin(x-3y)- 6y'sin(x-3)=0
sin(x-3y)-3y'sin(x-3y)=0
sin(x-3y)=3y'sin(x-3y)=0

sin(x-3y)/3=y'sin(x-3y)=0

sin(x-3y)
-------- = y'
3sin(x-3y)

1/3=y'

I think. I kinda didn't have time to hink about it.

come from?

i have to do that eventually?

*ducks*

derivative of sin^2 is actually 2sincos. Makes a big difference.

quote:

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•Delidgamond• had this to say about Punky Brewster:
take the derivitive of sin^2(x-3y)

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sin²(x-3y)

2cos(x-3y)*(1-3y)*(dy/dx)

Hell if I know, I hate math. I will never use this shit in real life.

quote:

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How.... •Delidgamond•.... uughhhhhh:
take the derivitive of sin^2(x-3y)
--------------
2sin(x-3y) * (1-3y`) = dx/dy

y` is y prime

I don't think that is right...the dx/dy is what is throwing me off...

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Now, back to work!

2sincos(x-3y) * (1-3y')=0

2sincos(x-3y) - 6y'sincos(x-3y)=0

2sincos(you)=6y'sincos(x-3y)

2sincos(that thingie)
------------------- =y'
6sincos(thatthingie)

1sincos(x-3y)
-------------------- =y'
3sincos(x-3y

1/3=y'