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

1. Jun 3, 2012

### rectifryer

1. The problem statement, all variables and given/known data

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

2. Relevant equations
None

3. 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.

2. Jun 3, 2012

### Villyer

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?

3. Jun 3, 2012

### Staff: Mentor

Well the relevant equation under 2. Relevant 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.

4. Jun 3, 2012

### rectifryer

That doesn't really seem like it would get me anywhere. I know I am wrong, but why would that work?

Last edited: Jun 3, 2012
5. Jun 3, 2012

### rectifryer

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 whats required to answer.

6. Jun 3, 2012

### HallsofIvy

Staff Emeritus
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 $\partial s/\partial y$?

Last edited by a moderator: Jun 3, 2012