Partial Derivatives
Partial derivatives
Let be a function of two variables. We define the partial derivative of with respect to as follows:
It means that it is the rate of change of with respect to when is fixed.
Similarly, the partial derivative of with respect to is as follows:
It means that it is the rate of change of with respect to when is fixed.
Suppose we find the values of the partial derivatives of at point . We can use the one of the following notations:
All the differentiation rules for ordinary derivatives can be used to compute partial derivatives:
To compute , is treated as a constant and differentiate with respect to .
To compute , is treated as a constant and differentiate with respect to .
Examples:
Let . Find and .
Answer:
The partial derivatives for functions of three variables are similarly defined. For example, let be a function of three variables. Then we have
i.e. it is the rate of change of with respect to while both and are kept fixed.
Example:
Let . Find and .
Answer:
The following applet, the surface is the graph of . The red and green curves on the graph are the curves cut out by the vertical plane through the point in the direction of x-axis and y-axis respectively. And the partial derivatives with respect to and can be interpreted as the slope of the red and green curves i.e the slope of the graph in x-direction and y-direction respectively.
Exercise: Let . Find and .
Exercise: Let . Find the slope of the graph of in x-direction at .
Implicit partial differentiation
The technique of implicit differentiation can also be applied to partial derivatives. For example, consider the equation of sphere . We would like to find the slope of the sphere in x-direction at the point . As the point is on the upper hemisphere, we can rewrite the equation into the following:
Then the required slope of the sphere is . The calculation is as follows:
Hence,
Alternatively, we can use implicit differentiation as follows: Regarding as an implicit function of and i.e. , we differentiate the equation of sphere with respect to and we have
Partial derivatives and continuity
For functions of one variable, it is well known that if a function is differentiable at a point, then it is continuous at the point. However, for multivariable functions, the existence of partial derivatives does not guarantee the continuity of the function. The following is one such example:
First we find the partial derivatives of at :
i.e. both partial derivatives at exist. However, is not continuous at because we already know that does not exist! (See the example in "limit along a curve")
In view of this example, we need a notion of differentiability of multivariable functions which is stronger than the existence of partial derivatives. We will talk about differentiability later.
Higher order partial derivatives
For a function , and are also functions of and . Therefore, we can take partial derivatives of them as follows:
(Note: and are sometimes called the mixed partial derivatives.)
Example:
Let . Find and .
Answer:
Notice that in the above example. In fact, it can be shown that if both and are continuous, then . In other words, for functions that are "nice" enough, the order of taking mixed partial derivatives does not matter.
Remark: For functions of three variables, similar notations and definitions apply. For example, suppose is function of three variables. Then .
Exercise: Let . Find all second-order partial derivatives of .
Differentiability
Let us recall that for a function of one variable, say , we say is differentiable at if the following limit exists:
We can rewrite it as follows:
where , which can be regarded as the "error" in the linear approximation of by (the approximation using the tangent line at ). In other words, is differentiable at if such error goes to faster than goes to .
For a function of two variables , we can use the same idea to define what it means to say is differentiable at .
In the applet below, we consider and to be the small change from . Then we use the tangent plane at to the graph of to approximate . It can be shown that linear approximation of is
(Note: In the applet below, and are abbreviated as and respectively)
We can define the error of linear approximation as follows:
Therefore, we have the following definition:
Definition: is differentiable at if i.e.
Remark: In the above definition, for to be differentiable, the existence of its partial derivatives is required. But we will see that the existence of partial derivatives cannot guarantee that the function is differentiable.
Example: Suppose . Prove that is differentiable at .
Answer:
First of all, and .
We need to compute the following limit:
Example: Suppose . Prove that is not differentiable at .
Answer:
First of all, we compute :
It is well known that such limit does not exist. Therefore, does not exist and hence is not differentiable at .
Example: Suppose . Show that its partial derivatives exist at but it is not differentiable at .
Answer:
First of all, we compute and :
Hence, both partial derivatives exist.
To check whether is differentiable at , we consider the following limit:
Consider the curve : as . Then
Consider the curve : as . Then
Hence, the limit does not exist and is not differentiable at .
(Note: You can use GeoGebra to plot the graph of this function to see why there is no tangent plane to the graph at .)
Exercise: Let . Prove the it is differentiable at but not differentiable at .
Theorems about differentiability
It is usually quite tedious to check whether a function of two variables is differentiable at a point using the above definition. The following is a theorem which gives a sufficient condition for a function of two variables to be differentiable at a point:
Theorem: Let be a function such that all its partial derivatives and exist and continuous at , then is differentiable at .
Example: Suppose . Show that is differentiable everywhere.
Answer:
Since polynomial functions and trigonometric functions are continuous, and are continuous everywhere.
By the above theorem, is differentiable everywhere.
The following theorem says that differentiability is a stronger condition than continuity:
Theorem: If is differentiable at , then is continuous at .
Proof:
It suffice to verify that .
Let and . We can rewrite the above limit and the condition becomes as follows:
Since is differentiable at , we have
Then we compute the following limit:
(since all the terms tend to zero)
This completes the proof.
Remarks:
- The contrapositive of this theorem is useful - if a function is not continuous at a point, then it is not differentiable at the point.
- The converse of the above theorem is certainly false. Consider the previous example . It can be shown that it is continuous at (left as exercise). However, we have already shown that it is not differentiable at .