1. Section 11.3 Partial Derivatives
Goals
Define partial derivatives
Learn notation and rules for calculating partial derivatives
Interpret partial derivatives
Discuss higher derivatives
Apply to partial differential equations

2. Introduction
If f is a function of two variables x and y, suppose we let only x vary while keeping y fixed, say y = b, where b is a constant.
Then we are really considering a function of a single variable x, namely, g(x) = f(x, b).
If g has a derivative at a, then we call it the partial derivative of f with respect to x at (a, b) and denote it by fx(a, b).

3. Introduction (cont’d)
Thus
The definition of a derivative gives

4. Introduction (cont’d)
Similarly, the partial derivative of f with respect to y at (a, b), denoted by fy(a, b), is obtained by holding x = a and finding the ordinary derivative at b of the function G(y) = f(a, y):

5. Definition
If we now let the point (a, b) vary, fx and fy become functions of two variables:

6. Notations
There are many alternate notations for partial derivatives:

7. Finding Partial Derivatives
The partial derivative with respect to x is just the ordinary derivative of the function g of a single variable that we get by keeping y fixed:

8. Example If f(x, y) = x3 + x2y3 – 2y2, find fx(2, 1) and fy(2, 1).
Solution Holding y constant and differentiating with respect to x, we get
fx(x, y) = 3x2 + 2xy3
and so
fx(2, 1) = 3 · 22 + 2 · 2 · 13 = 16

9. Solution (cont’d)
Holding x constant and differentiating with respect to y, we get
fx(x, y) = 3x2y2 – 4y
and so
fx(2, 1) = 3 · 22 · 12 – 4 · 1 = 8

10. Interpretations
To give a geometric interpretation of partial derivatives, we recall that the equation z = f(x, y) represents a surface S.
If f(a, b) = c, then the point P(a, b, c) lies on S. By fixing y = b, we are restricting our attention to the curve C1 in which the vertical plane y = b intersects S.

11. Interpretations (cont’d)
Likewise, the vertical plane x = a intersects S in a curve C2.
Both of the curves C1 and C2 pass through the point P.
This is illustrated on the next slide:

12. Interpretations (cont’d)

13. Interpretations (cont’d)
Notice that…
C1 is the graph of the function g(x) = f(x, b), so the slope of its tangent T1 at P is g′(a) = fx(a, b);
C2 is the graph of the function G(y) = f(a, y), so the slope of its tangent T2 at P is G′(b) = fy(a, b).
Thus fx and fy are the slopes of the tangent lines at P(a, b, c) to C1 and C2.

14. Interpretations (cont’d)
Partial derivatives can also be interpreted as rates of change.
If z = f(x, y), then…
∂z/∂x represents the rate of change of z with respect to x when y is fixed.
Similarly, ∂z/∂y represents the rate of change of z with respect to y when x is fixed.

15. Example If f(x, y) = 4 – x2 – 2y2, find fx(1, 1) and fy(1, 1).
Interpret these numbers as slopes.
Solution We have

16. Solution (cont’d)
The graph of f is the paraboloid z = 4 – x2 – 2y2 and the vertical plane y = 1 intersects it in the parabola z = 2 – x2, y = 1.
The slope of the tangent line to this parabola at the point (1, 1, 1) is fx(1, 1) = –2.
Similarly, the plane x = 1 intersects the graph of f in the parabola z = 3 – 2y2, x = 1.

17. Solution (cont’d)
The slope of the tangent line to this parabola at the point (1, 1, 1) is fy(1, 1) = –4:

18. Example If
Solution Using the Chain Rule for functions of one variable, we have

19. Example
Find ∂z/∂x and ∂z/∂y if z is defined implicitly as a function of x and y by
x3 + y3 + z3 + 6xyz = 1
Solution To find ∂z/∂x, we differentiate implicitly with respect to x, being careful to treat y as a constant:

20. Solution (cont’d) Solving this equation for ∂z/∂x, we obtain
Similarly, implicit differentiation with respect to y gives

21. More Than Two Variables
Partial derivatives can also be defined for functions of three or more variables, for example
If w = f(x, y, z), then fx = ∂w/∂x is the rate of change of w with respect to x when y and z are held fixed.

22. More Than Two Variables (cont’d)
But we can’t interpret fx geometrically because the graph of f lies in four-dimensional space.
In general, if u = f(x1, x2 ,…, xn), then
and we also write

23. fy = xexy ln z and fz = exy/z
Example
Find fx , fy , and fz if f(x, y, z) = exy ln z.
Solution Holding y and z constant and differentiating with respect to x, we have
fx = yexy ln z
Similarly,
fy = xexy ln z and fz = exy/z

24. (fx)x , (fx)y , (fy)x , and (fy)y ,
Higher Derivatives
If f is a function of two variables, then its partial derivatives fx and fy are also functions of two variables, so we can consider their partial derivatives
(fx)x , (fx)y , (fy)x , and (fy)y ,
which are called the second partial derivatives of f.

25. Higher Derivatives (cont’d)
If z = f(x, y), we use the following notation:

26. fx(x, y) = 3x2 + 2xy3 fy(x, y) = 3x2y2 – 4y
Example
Find the second partial derivatives of
f(x, y) = x3 + x2y3 – 2y2
Solution Earlier we found that
fx(x, y) = 3x2 + 2xy fy(x, y) = 3x2y2 – 4y
Therefore

27. Mixed Partial Derivatives
Note that fxy = fyx in the preceding example, which is not just a coincidence.
It turns out that fxy = fyx for most functions that one meets in practice:

28. Mixed Partial Derivatives (cont’d)
Partial derivatives of order 3 or higher can also be defined. For instance,
and using Clairaut’s Theorem we can show that fxyy = fyxy = fyyx if these functions are continuous.

29. Example
Calculate fxxyz if f(x, y, z) = sin(3x + yz).
Solution

30. Partial Differential Equations
Partial derivatives occur in partial differential equations that express certain physical laws.
For instance, the partial differential equation
is called Laplace’s equation.

31. Partial Differential Eqns. (cont’d)
Solutions of this equation are called harmonic functions and play a role in problems of heat conduction, fluid flow, and electric potential.
For example, we can show that the function u(x, y) = ex sin y is a solution of Laplace’s equation:

32. Partial Differential Eqns. (cont’d)
Therefore, u satisfies Laplace’s equation.

33. Partial Differential Eqns. (cont’d)
The wave equation
describes the motion of a waveform, which could be an ocean wave, sound wave, light wave, or wave traveling along a string.

34. Partial Differential Eqns. (cont’d)
For instance, if u(x, t) represents the disaplacement of a violin string at time t and at a distance x from one end of the string, then u(x, t) satisfies the wave equation. (See the next slide.)
Here the constant a depends on the density of the string and on the tension in the string.

35. Partial Differential Eqns. (cont’d)

36. Example
Verify that the function u(x, t) = sin(x – at) satisfies the wave equation.
Solution Calculation gives
So u satisfies the wave equation.

37. Review Definition of partial derivative
Notations for partial derivatives
Finding partial derivatives
Interpretations of partial derivatives
Function of more than two variables
Higher derivatives
Partial differential equations