Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.


Variations in $x,y,z$ and $X$ at constant $t$ are independent of $t$ (since each of these variables is strictly a function of $t$), so we have $${\frac{\partial x}{\partial X}=\frac{\partial{\dot x}}{\partial{\dot X}}}$$

(This is just after equation (5) on the page.)

I'm having trouble making sense of this. If each of these variables is strictly a function of $t$, and $t$ is held constant, how does a time derivative (${\dot x}$) make sense?

If it were to mean that $t$ is replaced with a constant after the differentiation, then I could take the example of

$$ x(y)=X^{3/4}=t^{3}$$ $$ X(t)=t^{4}$$

I could then calculate

$$ {\frac{\partial x}{\partial X}=\frac{\dot x}{\dot X} = \frac{3}{4t}}$$


$$ {\frac{\partial {\dot x}}{\partial {\dot X}}=\frac{\ddot x}{\ddot X}=\frac{1}{2t}}$$

...and those are not equal for any $t$.

What is the justification for this step?

Note: This is a follow up to my first question about this, but that error was resolved. Rates of change for functions dependent on same variable

share|cite|improve this question

1 Answer 1

up vote 1 down vote accepted
  • Between eq. (4) and eq. (5)

By definition


and not $x=x(X(t),Y(t),t)$; so

$$\frac{dx}{dt}=\frac{\partial x}{\partial X}\frac{dX}{dt}+ \frac{\partial x}{\partial Y}\frac{dY}{dt},$$ from which follows $$\frac{\partial }{\partial X}\left(\frac{dx}{dt}\right)= \frac{\partial^2 x}{\partial X^2}\frac{dX}{dt}+ \frac{\partial^2 x}{\partial X\partial Y}\frac{dY}{dt},$$

as $\frac{\partial }{\partial X}\frac{dX}{dt}=\frac{\partial }{\partial X}\frac{dY}{dt}=0.$ On the other hand

$$\frac{d}{dt}\left(\frac{\partial x}{\partial X}\right):= \frac{d}{dt}g(X(t),Y(t))= \frac{\partial g}{\partial X}\frac{dX}{dt}+ \frac{\partial g}{\partial Y}\frac{dY}{dt} = \frac{\partial^2 x}{\partial X^2}\frac{dX}{dt}+ \frac{\partial^2 x}{\partial X\partial Y}\frac{dY}{dt}, $$


$$\frac{\partial }{\partial X}\left(\frac{dx}{dt}\right) =\frac{d}{dt}\left(\frac{\partial x}{\partial X}\right).$$

  • Just after eq. (5)

As $$x=x(X(t),Y(t)),$$ then $\frac{\partial x }{\partial \dot{X}}=0$ and $\frac{\partial x }{\partial \dot{X}}\left(\frac{\partial x}{\partial X} \right)=0,$ as well. Using the notation $\dot{x}=\frac{dx}{dt} $ it follows that

$$\frac{\partial\dot{x} }{\partial \dot{X}}= \frac{\partial }{\partial \dot{X}}\left(\frac{\partial x}{\partial X}\dot{X}+ \frac{\partial x}{\partial Y}\frac{dY}{dt},\right)=\frac{\partial x}{\partial X},$$

as claimed.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.