Total differentiation of scalar fields

Total differentiation of scalar fields

Total differentiation

Consider a multivariate function.

\(f(x)\).

We can define:

\(\Delta f(x, \Delta x):=f(x+\Delta x)-f(x)\)

\(\Delta f(x, \Delta x)=\sum_{i=1}^nf(x+\Delta x_i+\sum_{j=0}^{i-1}\Delta x_j)-f(x+\sum_{j=0}^{i-1}\Delta x_j)\)

\(\Delta f(x, \Delta x)=\sum_{i=1}^n\Delta x_i \dfrac{f(x+\Delta x_i+\sum_{j=0}^{i-1}\Delta x_j)-f(x+\sum_{j=0}^{i-1}\Delta x_j)}{\Delta x_i}\)

\(\dfrac{\Delta f}{\Delta x_k}=\sum_{i=1}^n\dfrac{\Delta x_i}{\Delta x_k} \dfrac{f(x+\Delta x_i+\sum_{j=0}^{i-1}\Delta x_j)-f(x+\sum_{j=0}^{i-1}\Delta x_j)}{\Delta x_i}\)

\(\lim_{\Delta x_k \rightarrow 0}\dfrac{\Delta f}{\Delta x_k}=\sum_{i=1}^n\lim_{\Delta x_k \rightarrow 0}\dfrac{\Delta x_i}{\Delta x_k} \dfrac{f(x+\Delta x_i+\sum_{j=0}^{i-1}\Delta x_j)-f(x+\sum_{j=0}^{i-1}\Delta x_j)}{\Delta x_i}\)

\(\dfrac{df}{dx_k}=\sum_{i=1}^n\dfrac{dx_i}{dx_k} \dfrac{\delta f}{\delta x_i}\)

Total differentiation of a univariate function

For a univariate function total differentiation is the same as partial differentiation.

\(\dfrac{df}{dx}=\dfrac{dx}{dx} \dfrac{\delta f}{\delta x}\)

\(\dfrac{df}{dx}=\dfrac{\delta f}{\delta x}\)