Suppose \({\bf c}\colon\, {\mathbb R}\rightarrow {\mathbb R}^3\) is a differentiable path and \(f\colon\, {\mathbb R}^3\rightarrow {\mathbb R}\) is a differentiable function. Let \(h(t)=f({\bf c}(t))=f(x(t),y(t),z(t)),\) where \({\bf c}(t)=(x(t),y(t),z(t))\). Then \begin{equation*} \frac\it dh\it dt= \frac{\partial f}{\partial x}\frac\it dx\it dt +\frac{\partial f}{\partial y}\frac\it dy\it dt +\frac{\partial f}{\partial z}\frac\it dz\it dt.\tag{2} \end{equation*}
Suppose \({\bf c}\colon\, {\mathbb R}\rightarrow {\mathbb R}^3\) is a differentiable path and \(f\colon\, {\mathbb R}^3\rightarrow {\mathbb R}\) is a differentiable function. Let \(h(t)=f({\bf c}(t))=f(x(t),y(t),z(t)),\) where \({\bf c}(t)=(x(t),y(t),z(t))\). Then \begin{equation*} \frac\it dh\it dt= \frac{\partial f}{\partial x}\frac\it dx\it dt +\frac{\partial f}{\partial y}\frac\it dy\it dt +\frac{\partial f}{\partial z}\frac\it dz\it dt.\tag{2} \end{equation*}
Suppose \({\bf c}\colon\, {\mathbb R}\rightarrow {\mathbb R}^3\) is a differentiable path and \(f\colon\, {\mathbb R}^3\rightarrow {\mathbb R}\) is a differentiable function. Let \(h(t)=f({\bf c}(t))=f(x(t),y(t),z(t)),\) where \({\bf c}(t)=(x(t),y(t),z(t))\). Then \begin{equation*} \frac\it dh\it dt= \frac{\partial f}{\partial x}\frac\it dx\it dt +\frac{\partial f}{\partial y}\frac\it dy\it dt +\frac{\partial f}{\partial z}\frac\it dz\it dt.\tag{2} \end{equation*}