This picture can be used to "see" the relationship ofto dy ; recall that = dx In the picture x = c is the "starting point" , so to speak , and the functions values at c and c +  have a difference . 

= f ( c + ) - f( c ) Recall that dy = f ' (c) by definition or dy = f '(c) dx . Find and dy in the picture. Notice that dy is to the tangent line instead of to the curve and that , for differentiable functions , the difference between the and the dy is "squeezed" to zero, so to speak , as -->0 .

This idea is presented in Calculus I but may very well be lost or not understood at the time. I bring it to your attention because it is the " way into " the idea of differentiability of functionsof more than one independent variable.

With = f '( c ) dx + and -->0 as -->0 in mind  , we proceed to "see" the same idea in the following:

For z = f ( x,y ) = () dx + () dy+ dx + dy and --> 0 as -->0 and-->0 as -->0 if f (x) is a differentiable function in x and y at a particular point and . I'm assuming the partial derivatives in this equation are computed at the particular point.

The bold underlined item in the equation is to the formula for what f '( c ) dx is to the formula for

above it.

In the first case we have the derivative multiplied by dx , in the second case we have the sum of the partial derivatives multiplied by the corresponding dx and dy , and if there are more independent variables we will have more products to add.

In the first case the f '( c ) dx is the differential  and in the second case dx + dy is the differential and is often referred to as the total differential (emphasis on more than one independent variable) .

Look at the following solved problems in the Exercises at the end of 13.4 : 1-11 odd Then
do the following unsolved problems as homework: 2,4,8,10,12