1. Sep 20, 2004

da_willem

Suppose you have a function x(t)=ct^2. Then you are asked to compute dx'/dx, where x' denotes a derivative to time; x'=dx/dt.

I: $$\frac{dx'}{dx}=\frac{d}{dx} \frac{dx}{dt} =\frac{d}{dt} \frac{dx}{dx}= \frac{d}{dt}(1)=0$$ (By using Clairaut's theorem)

II: $$x'=2ct=2c \sqrt{\frac{x}{c}}=2 \sqrt{cx} -> \frac{dx'}{dx}=\frac{d}{dx} 2 \sqrt{cx} = \sqrt{\frac{c}{x}}$$

Can someone clarify this for me?

2. Sep 20, 2004

matt grime

x and t aren't "independent"

dy/dx =(dy/dz)(dz/dx)

doesn't work for this kind situation.

3. Sep 20, 2004

da_willem

Why would the variables have to be independent? And I don't believe to have used the chain rule in any crucial part anyway, so could you please explain what's wrong

4. Sep 20, 2004

matt grime

x and t are implicitly defined as functions of each other by the equation x-ct^2=0.

so they are not necessarily going to have mixed partial derivatives that are equal, since they are not independent variables..

sorry for the OT bit on the chain rule.

5. Sep 20, 2004

da_willem

So dx(t)'/dx is not necessarily zero if t and x are dependent because then Clairaut's theorem (the equality of the mixed partial derivatives) does not apply. So in the example the second line of reasoning is correct while the first one is flawed?

6. Sep 20, 2004

matt grime

yep.

what does clariat say? that for suitable f that:

f_{xy} = f_{yx} well, what's the f that x and t are both functions of?
clairaut just doesn't apply here since there is no function of two (or more) variables lying around to differentiate)

7. Sep 20, 2004

da_willem

k, thank you very much. To me it is clear now.

8. Sep 20, 2004

krab

One thing you CAN say is
$${d\over dx}x'={dt\over dx}{d\over dt}x'=x''/x'$$
In your example, this is 1/t.

Last edited: Sep 20, 2004