Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/219

We shall next consider a function of two variables, both of which depend on a single independent variable. Consider the function

$u=f(x,\ y),$

where $x$ and $y$ are functions of a third variable $t$.

Let $t$ take on the increment $\Delta t$, and let $\Delta x$, $\Delta y$, $\Delta u$ be the corresponding increments of $x$, $y$, $u$ respectively. Then the quantity

$\Delta u=f(x+\Delta x,\ y+\Delta y)-f(x,\ y)$

is called the total increment of $u$.

Adding and subtracting $f(x,y+\Delta y)$ in the second member,

(A)	$\Delta u=[f(x+\Delta x,\ y+\Delta y)-f(x,\ y+\Delta y)]+[f(x,\ y+\Delta y)-f(x,\ y)].$

Applying the Theorem of Mean Value (46), §106, to each of the two differences on the right-hand side of (A), we get, for the first difference,

(B)	$f(x+\Delta x,y+\Delta y)-f(x,y+\Delta y)=f_{x}'(x+\theta _{1}\cdot \Delta x,y+\Delta y)\Delta x.$
	[$a=x,\Delta a=\Delta x$, and since $x$ varies while $y+\Delta y$ remains constant, we get the partial derivative with respect to $x$.]

For the second difference we get

(C)	$f(x,y+\Delta y)-f(x,y)=f_{y}'(x,y+\theta _{2}\cdot \Delta y)\Delta y.$
	[$a=y,\delta a=\Delta y,$ and since $y$ varies while $x$ remains constant, we get the partial derivative with respect to $y$.]

Substituting (B) and (C) in (A) gives

(D)	$\Delta u=f_{x}'(x+\theta _{1}\cdot \Delta x,y+\Delta y)\Delta x+f_{y}'(x,y+\theta _{2}\cdot \Delta y)\Delta y,$

where $\theta _{1}$ and $\theta _{2}$ are positive proper fractions. Dividing (D) by $\Delta t$,

(E)	${\frac {\Delta u}{\Delta t}}=f_{x}'(x+\theta _{1}\cdot \Delta x,y+\Delta y){\frac {\Delta x}{\Delta t}}+f_{y}'(x,y+\theta _{2}\cdot \Delta y){\frac {\Delta y}{\Delta t}}.$

Now let $\Delta t$ approach zero as a limit, then

(F)	${\frac {du}{dt}}=f_{x}'(x,y){\frac {dx}{dt}}+f_{y}'(x,y){\frac {dy}{dt}}.$
	[Since $\Delta x$ and $\Delta y$ converge to zero with $\Delta t$, we get $\lim _{\Delta t\to 0}f_{x}'(x+\theta _{1}\cdot x,y+\Delta y)=f_{x}'(x,y),$ and $\lim _{\Delta t\to 0}f_{y}'(x,y+\theta _{2}\cdot \Delta y)=f_{y}'(x,y),f_{x}'(x,y)$ and $f_{y}'(x,y)$ being assumed continuous.]

Replacing $f(x,y)$ by $u$ in (F), we get the total derivative

(51)	${\frac {du}{dt}}={\frac {\partial u}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial u}{\partial y}}{\frac {dy}{dt}}.$