Need help with chain rule for relating ds/dt to dx/dt and dy/dt

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Homework Statement



s=\sqrt{(3x^2)+(6y^2)}

Homework Equations


None


The Attempt at a Solution


\stackrel{ds}{dt}=\stackrel{d}{dt}\sqrt{(3x^2)+(6y^2)}

\stackrel{3x}{\sqrt{(3x^2)+(6y^2)}}

The problem with that is its only d/dx if y is a set number. I don't know how to differentiate the entire thing properly. I have been hacking at this for 8 hours. I feel like mental jello.
 
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You are taking the derivative with respect to t.

So d/dt of 3x2 = 6x * dx/dt, not 6x.Maybe this helps figure out the whole derivative?
 
Well the relevant equation under 2. Homework Equations would be an expression of the chain rule.

d/dt(f(g(t)) = f'(g(t))*g'(t)

http://archives.math.utk.edu/visual.calculus/2/chain_rule.4/index.html
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html
http://mathworld.wolfram.com/ChainRule.html

Let g(t) = g(x(t),y(t)) and f = √

One could also write the original equations as s2 = 3x2 + 6y2, and differentiate each term with respect to t.
 
One could also write the original equations as s2 = 3x2 + 6y2, and differentiate each term with respect to t.
That doesn't really seem like it would get me anywhere. I know I am wrong, but why would that work?

Thank you for the links.
 
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I have narrowed down my question, specifically to the area I have highlighted on this picture (bear in mind, I can't post pics under 10 posts):

http:// i.imgur .com /62erw.png

Where did all the dx/dt and dy/dt come from on the right side? I don't understand that step. I know how to do this when thinking about it in function form, but it confuses me to think about it in fraction form, which is what's required to answer.
 
If s is a function of two variables, x and y, which are themselves functions of t. The "chain rule" says
\frac{ds}{dt}= \frac{\partial s}{\partial x}\frac{dx}{dt}+ \frac{\partial s}{\partial y}\frac{dy}{dt}

Here, s(x,y)= \sqrt{3x^2+ 6y^2}= (3x^2+ 6y^2)^{1/2}
What are \partial s/\partial x and \partial s/\partial y?
 
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