The actual meaning of $\frac{\partial}{\partial{x}}$

calculusmultivariable-calculus

When I searched total derivative on wikipedia it says:

For $L=L(t,x_1(t),x_2(t),x_3(t)…x_n(t))$ the total derivative is given by:
$$\dfrac{\rm{d}L}{\rm{d}t}=\dfrac{\partial L}{\partial t}+\sum_i^n\dfrac{\partial L}{\partial x_i}\dfrac{\partial x_i}{\partial t}~.$$
So here $\dfrac{\partial L}{\partial t}$ is derivative of $L$ w.r.t an explicit independent variable $t$.

However, when I look for generalized chain rule, wikipedia says:

For $$y=y\left(u_1(x_1,x_2,…,x_i),u_2(x_1,x_2,…,x_i),…,u_m(x_1,x_2,…,x_i)\right)$$
the chain rule is given by: $$\frac{\partial y}{\partial x_i}=\sum_{l=1}^m\dfrac{\partial y}{\partial u_l}\dfrac{\partial u_l}{\partial
x_i}$$

Now the partial derivative operator $\dfrac{\partial}{\partial{x_i}}$ looks more like a total derivative, rather than a derivative w.r.t an explicit independent variable.

Could anybody tell me what on earth is $\dfrac{\partial}{\partial{x_i}}$ ?

Thanks!

Best Answer

In the second expression $\frac{\partial y}{\partial x_i}$ is in fact a partial derivative too, and this is because $y$ is ultimately a function of many variables $x_1,\ldots,x_n$. In your first expression $\frac{dL}{dt}$ a total derivative is possible here because the function is ultimately a function of a single variable, namely $t$. This is the main reason why a total derivative for $y$ wouldn’t be possible in this case: $L$ ultimately is a function of a single variable while $y$ is a multivariable function.

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