Let $f:\mathbb R^n\rightarrow \mathbb R^m$ be a function. By definition, $f$ is differentiable at $a$ if there exists a linear map $D_af:\mathbb R^n\rightarrow\mathbb R^m$ such that
$$\lim_{h\rightarrow 0}\dfrac{||f(a+h)-f(a)-D_af(h)||}{||h||}=0$$.
When this limit exists, we call $D_af$ the total derivative of $f$ at $a$ and we call the corresponding matrix with respect to usual basis, the Jacobian matrix $J_af$. Now $f$ can be written as $f=(f_1,\cdots,f_m)$ where each $f_i:\mathbb R^n\rightarrow \mathbb R$ and $f(x)=(f_1(x),\cdots,f_m(x))$.
We define the partial derivative of $f_i$ at $a=(a_1,\cdots,a_n)$ in the direction of $x_j$ by the real number (if it exists)
$$\dfrac{\partial f_i}{\partial x_j}(a)=\lim_{h\rightarrow 0}\dfrac{f_i(a_1,\cdots,a_j+h,\cdots,a_n)-f_i(a)}{h}$$
And we show that when $f$ is differentiable at $a$ then all the real numbers $\dfrac{\partial f_i}{\partial x_j}(a)$ exist and the matrix elements $(J_af)_{i,j}=\dfrac{\partial f_i}{\partial x_j}(a)$.
Now I see in wikipedia articles that they give the name total derivative also for some other notion: given a map $g:\mathbb R^n\rightarrow \mathbb R$ then the total derivative of $g$ with respect to $x_j$ is
$$\dfrac{dg}{dx_j}=\dfrac{\partial g}{\partial x_1}\dfrac{dx_1}{dx_j}+\cdots+\dfrac{\partial g}{\partial x_n}\dfrac{dx_n}{dx_j}$$
My questions:
1) Is the formula above a definition for $\dfrac{dg}{dx_j}$ ?
2) Is the notion of $\dfrac{dg}{dx_j}$ reserved only for real valued maps $g:\mathbb R^n\rightarrow \mathbb R$ since it has partial derivatives in its definition formula?
3) How does these two notions of total derivative relate?
Best Answer
The total derivative is only a piece of notation to overcome some difficulties when dealing with Leibniz notation.
Consider the function $f(x,y)=x^2+y$. If you agree that $x$ and $y$ in the definition of $f$ are just placeholders you should agree that $f(y,x)=y^2+x$ and that $f(t,t^2)=2t^2$.
Now the question is: what does $\partial f / \partial x$ means? Written like that one would interpret $\partial/\partial x$ as the derivative of $f$ with respect to the first variable. Notice here that the variable $x$ is no more a simple place-holder but has a conventional meaning.
For example consider the following: $$ \frac{\partial f(y,x)}{\partial x}. $$ Now the interpretation is not clear... you mean the derivative with respect to the first or the second variable?
Mixing variables like that is not good... but sometimes one should be prepared to solve the ambiguity. Consider a function $f(x,t)$ which represent a quantity which depends on space $x$ and time $t$. So it is understood that $\partial f/\partial t$ is the derivative with respect to the second component (which is time). Suppose now that you have a particle which moves with the law $x=t^2$. If you evaluate the function $f$ on the particle you get $$ f(x(t),t) $$ and if you want to compute the derivative of this function you can use the chain rule and obtain: $$ \frac{d}{dt} f(x(t),t) = \frac{\partial}{\partial x} f(x(t),t)\cdot x'(t) + \frac{\partial }{\partial t} f(x(t),t). $$ Now the point is that often it is useful to reduce the notation writing $x$ in place of $x(t)$ and $f$ in place of $f(x,t)$ so that previous formula could be written as $$ \frac{d}{dt} f(x,t) = \frac{\partial}{\partial x} f(x,t) \cdot x'(t) + \frac{\partial }{\partial t} f(x,t) $$ or $$ \frac{df}{dt} = \frac{\partial f}{\partial x} \cdot x' + \frac{\partial f}{\partial t}. $$
Now you see that $d/dt$ and $\partial/\partial t$ assume different meanings...