If a variable, z, is a function of two variables, x and y, then to visualise it graphically we need not just a 2-dimensional curve but a 3-dimensional surface. 

Here is the surface representing the function z = x2 +  2y2

The point shown in blue is at x = -0.5, y = -0.5

You can see that the steepness of the function is different, not just at different points but in different directions at any one point.

The partial derivatives, ∂z/∂x and ∂z/∂y are functions that define the gradient of the tangent to the surface in the special cases where y is constant or x is constant, respectively. For example, the intersection with the plane defined by x = -0.5 is shown here:

The slice through the  z = x2 +  2y2 surface, created by this intersection with the x = -0.5 plane,  is a 2-dimensional curve, shown below:

Here is this tangent (shown in yellow), in the plane x= -0.5, shown with respect to the z = x2 +  2y2 surface: 

Now lets turn our attention to the other case, when we now hold y constant. Our point is at x = y = -0.5, let's look at the intersection of the z = x2 +  2y2 surface with the plane defined by y = -0.5:

And here is the curve defined by the intersection of this surface with the y = -0.5 plane:

And here is this tangent (shown in yellow), in the plane y = -0.5, shown with respect to the z = x2 +  2y2 surface: 

These two tangents obviously intersect at the point x = -0.5 and y = -0.5.  Like any two intersecting straight lines, the two tangents define a plane. This is the tangent plane to the z = x2 +  2y2 surface:

So the partial derivatives give us the slope of this tangent plane in two specific directions: constant x and constant y.  In general, though, we may want to know the slope in any direction.  How could we find this?

In order to approach this problem, let's first make a simplifying change to our tangent plane.  We have found, in question 4A-3, that the equation of this plane is z = -x - 2y - 0.75.  If we differentiate this, with respect to either x or y, the constant makes no difference. This means that the slope of the plane is the same if we just drop the constant from the equation and focus on the plane z = -x - 2y. This will make our lives easier since it now passes through the origin of our coordinate system.  Here's a comparison of the two planes, just to convince you that the slope really is the same:

Now, if we use our new plane, z = -x- 2y, we can visualise more easily the meaning of the partial derivatives:

In the picture above, the red line represents the slope of the plane in the direction where y is constant. In other words, the gradient of this line is equal to ∂z/∂x. Similarly, the green line  represents the slope when x is constant, so the gradient of this line is ∂z/∂y. 

Suppose, however, we want to know the slope of the plane in a different direction. We'll take an example where the direction in the horizontal plane, z=0,  is x=y. In the picture below, this is the white line. The slope of our plane in this direction is therefore the gradient of the yellow line:

First of all, let's try and figure this out trigonometrically:

This question shows that finding the equation for the tangent plane, by using the partial derivatives, makes it possible to find the slope of this plane and hence of the original surface, in any direction.

However it would be nice if we could do this without having to wrestle with the tangent surface and get there directly from the partial derivatives.  We can do this using a vector approach:

We can generalise this method to any point on any surface in any direction:

Let's say the point at which the tangents are drawn is (x0, y0).

The direction in which the slope is to measured is defined by the combination of the unit vectors in the horizontal plane, mi + nj

Then the gradient is given by:      (m∂z/∂x  +  n∂z/∂y) / √(m2 +  n2)

(where ∂z/∂x and ∂z/∂y are the values at (x0, y0).