Section 14.3 Partial Derivatives
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1 Section 14.3 Partial Derivatives Ruipeng Shen March 20 1 Basic Conceptions If f(x, y) 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) f x (a, b) = g f(a + h, b) f(a, b) (a) = lim. Definition 1. If f is a function of two variables, its partial derivatives are the functions f x and f y defined by f(x + h, y) f(x, y) f(x, y + h) f(x, y) f x (x, y) = lim ; f y (x, y) = lim. There are many alternative notations for partial derivatives of z = f(x, y) as below f x (x, y) = f x = = f(x, y) = = f 1 = D 1 f = D x f f y (x, y) = f y = y = f(x, y) = y y = f 2 = D 2 f = D y f Rule for Finding Partial Derivatives 1. To find f x, regard y as a constant and differentiate f(x, y) with respect to x. 2. To find f y, regard x as a constant and differentiate f(x, y) with respect to y. Example 2. If f(x, y) = x 3 + sin(xy), find f x (1, 1) and f y (1, 1). Basic calculation shows f x = 3x 2 + y cos(xy); f y = x cos(xy). Therefore we have f x (1, 1) = 3 + cos 1; f y (1, 1) = cos 1. 1
2 z T 1 C 1 T 2 C 2 (a,b,0) y x Figure 1: Geometrical Interpretation of Partial Derivatives 2 Interpretations of Partial Derivatives The partial derivative f x (a, b) and f y (a, b) can be interpreted geometrically as the slopes of the tangent lines at P (a, b, f(a, b)) to the two traces C 1 and C 2 of S in the planes y = b and x = a. Example 3. Suppose that a differentiable function f(x, y) has a maximal value at (1, 3), as shown in the figure 2. What is the value of the partial derivative f x (1, 3) and f y (1, 3)? Since the tangent line at the peak is always horizontal, we know the slopes f x (1, 3) and f y (1, 3) must be zero. z (1,3,0) y x Figure 2: The maximum has horizontal tangent lines 2
3 3 More Differentiation Example 4. If f(x, y) = e x/y, calculate and y. Using the chain rule we have = ex/y y = ex/y ( ) x = 1 y y ex/y ; ( ) x y y = x y 2 ex/y. Example 5. Assume z is defined implicitly as a function of x and y by the equation Find the partial derivatives / and / y. x 3 + y 3 + z 3 + 6xyz = 1. Applying implicit differentiation with respect to x, we have Solving / from the identity above, we have 3x 2 + 3z 2 + 6yz + 6xy = 0 = + 2yz x2 z 2 + 2xy. In the same way we have y = + 2xz y2 z 2 + 2xy. 4 Higher Derivatives If z = f(x, y) is a function of two variables, then its partial derivatives f x and f y are also functions of two variables, so we can consider their partial derivatives (f x ) x, (f x ) y, (f y ) x and (f y ) y, which are called the second partial derivatives of f. We use the following notations (f x ) x = f xx = f 11 = ( ) (f x ) y = f xy = f 12 = ( ) y (f y ) x = f yx = f 21 = ( ) y (f y ) y = f yy = f 22 = ( ) y y 2 = 2 z 2 ; y = 2 z y ; y = 2 z y ; y 2 = 2 z y 2 ; Example 6. Find the second partial derivatives of f(x, y) = x 3 + x 2 y 3 2y 2. 3
4 We first calculate the first-order partial derivatives f x = 3x 2 + 2xy 3 ; f y = 3x 2 y 2 4y. Therefore we have f xx = ( 3x 2 + 2xy 3) = 6x + 2y 3 ; f xy = ( 3x 2 + 2xy 3) = 6xy 2 ; y f yx = ( 3x 2 y 2 4y ) = 6xy 2 ; f yy = ( 3x 2 y 2 4y ) = 6x 2 y 4; y Theorem 7 (Clairaut s Theorem). Suppose f is defined in a disk D containing the point (a, b). If the functions f xy and f yx are both continuous on D. then f xy (a, b) = f yx (a, b). We can also define derivatives of order 3 or higher. For instance f xyy = (f xy ) y = ( 2 ) f = 3 f y y y 2. Using Clairaut s Theorem, one can show f xyy = f yxy = f yyx as long as all partial derivatives of f are continuous. Example 8. Calculate f xxyz if f(x, y, z) = sin(3x + yz). Basic calculation shows f x = 3 cos(3x + yz); f xx = 9 sin(3x + yz); f xxy = 9z cos(3x + yz); f xxyz = 9 cos(3x + yz) + 9yz sin(3x + yz). 5 Partial Differential Equations Partial Differential Equations are equations that involve the partial derivatives of an unknown function. For example, the partial differential equation 2 u u y 2 = 0 is called Laplace s equation. s to this equation are called harmonic functions. Example 9. Show that the function u(x, y) = e x sin y is a solution to the Laplace s equation. By a basic calculation we have u x = e x sin y, u y = e x cos y; u xx = e x sin y, u yy = e y sin y; Therefore we have u xx + u yy = 0. Example 10. Let f be a smooth function. Verify the function u(x, t) = f(x at) satisfies the wave equation 2 u t 2 = a2 2 u 2. 4
5 By a basic calculation we have u t = af (x at) u tt = a 2 f (x at), u x = f (x at); u xx = f (x at); Therefore we have u tt = a 2 f (x at) = a 2 u xx = 0. 5
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