What variables does this partial derivative depend upon?

Question

If $u=u(x(t),t)$ is it the case that

$\displaystyle \frac{\partial u}{\partial x}(x(t),t)=g(t)$, i.e purely a function of $t$?

I don't think it is but my notes say otherwise

Answer 1

We see here a very common, but unfortunate notational sloppyness: The letter $x$ is used for two different things, namely

• as name of a certain function $t\mapsto x(t)$,
• as name for the first (independent) variable of the function $$u:\quad {\mathbb R}^2\to{\mathbb R},\qquad (x,t)\mapsto u(x,t)\ .$$

When one writes ${\partial u\over\partial x}$ one actually has "the partial derivative of $u$ with respect to its first variable", whatever its name, in mind. Therefore I'd rather write $(x,t)\mapsto u_{.1}(x,t)$ for this derivative, if the letter $x$ is used at the same time for other purposes. (Instead of $u_{.1}$ one could also write $\partial_1 u$ or similar.) After we have computed this $u_{.1}(\cdot,\cdot)$ we can plug in the given functions $t\mapsto x(t)$ and $t\mapsto t$ at the places of the cdots and so obtain the pullback $$g(t):=u_{.1}\bigl(x(t),t\bigr)\ ,$$ which is a function of the single real variable $t$.

Answer 2

if $u(x(t),t)=t(x(t))^2$, then $\partial_xu(x(t),t)=2tx(t)$, which is a function of $t$. If the variable $x$ was independent of $t$ and you had say $u(x,t) = tx^2$, the $\partial_x u=2xt$ is a function of both $x$ and $t$.

Answer 3

You could argue that it is purely a function of $t$, because $x(t)$ is purely a function of $t$. When you do a partial derivative, you have to substitute before doing the differentiation in order to obtain the correct answer.