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) fx (a, b) = g 0 (a) = lim h→0 f (a + h, b) − f (a, b) . h Definition 1. If f is a function of two variables, its partial derivatives are the functions fx and fy defined by f (x + h, y) − f (x, y) ; h→0 h fx (x, y) = lim f (x, y + h) − f (x, y) . h→0 h fy (x, y) = lim There are many alternative notations for partial derivatives of z = f (x, y) as below ∂f ∂ = f (x, y) = ∂x ∂x ∂f ∂ fy (x, y) = fy = = f (x, y) = ∂y ∂y fx (x, y) = fx = ∂z = f1 = D 1 f = D x f ∂x ∂z = f2 = D2 f = Dy f ∂y Rule for Finding Partial Derivatives 1. To find fx , regard y as a constant and differentiate f (x, y) with respect to x. 2. To find fy , regard x as a constant and differentiate f (x, y) with respect to y. Example 2. If f (x, y) = x3 + sin(xy), find fx (1, 1) and fy (1, 1). Solution Basic calculation shows fx = 3x2 + y cos(xy); fy = x cos(xy). Therefore we have fx (1, 1) = 3 + cos 1; fy (1, 1) = cos 1. 1 z T1 C1 T2 C2 y (a,b,0) x Figure 1: Geometrical Interpretation of Partial Derivatives 2 Interpretations of Partial Derivatives The partial derivative fx (a, b) and fy (a, b) can be interpreted geometrically as the slopes of the tangent lines at P (a, b, f (a, b)) to the two traces C1 and C2 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 fx (1, 3) and fy (1, 3)? Solution Since the tangent line at the “peak” is always horizontal, we know the slopes fx (1, 3) and fy (1, 3) must be zero. z (1,3,0) y x Figure 2: The maximum has horizontal tangent lines 2 3 More Differentiation Example 4. If f (x, y) = ex/y , calculate Solution ∂f ∂f and . ∂x ∂y Using the chain rule we have ∂f = ex/y · ∂x ∂f = ex/y · ∂y ∂ x = ∂x y ∂ x = ∂y y 1 x/y e ; y −x x/y e . y2 Example 5. Assume z is defined implicitly as a function of x and y by the equation x3 + y 3 + z 3 + 6xyz = 1. Find the partial derivatives ∂z/∂x and ∂z/∂y. Solution Applying implicit differentiation with respect to x, we have 3x2 + 3z 2 · ∂z ∂z + 6yz + 6xy · =0 ∂x ∂x Solving ∂z/∂x from the identity above, we have ∂z x2 + 2yz =− 2 . ∂x z + 2xy In the same way we have y 2 + 2xz ∂z =− 2 . ∂y z + 2xy 4 Higher Derivatives If z = f (x, y) 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 . We use the following notations ∂ ∂f ∂2f ∂2z (fx )x = fxx = f11 = = = ; 2 ∂x ∂x ∂x ∂x2 ∂ ∂f ∂2f ∂2z = = ; (fx )y = fxy = f12 = ∂y ∂x ∂y∂x ∂y∂x ∂ ∂f ∂2f ∂2z (fy )x = fyx = f21 = = = ; ∂x ∂y ∂x∂y ∂x∂y ∂ ∂f ∂2f ∂2z (fy )y = fyy = f22 = = = ; ∂y ∂y ∂y 2 ∂y 2 Example 6. Find the second partial derivatives of f (x, y) = x3 + x2 y 3 − 2y 2 . 3 Solution We first calculate the first-order partial derivatives fx = 3x2 + 2xy 3 ; fy = 3x2 y 2 − 4y. Therefore we have ∂ fxx = 3x2 + 2xy 3 = 6x + 2y 3 ; ∂x ∂ fyx = 3x2 y 2 − 4y = 6xy 2 ; ∂x ∂ 3x2 + 2xy 3 = 6xy 2 ; ∂y ∂ = 3x2 y 2 − 4y = 6x2 y − 4; ∂y fxy = fyy Theorem 7 (Clairaut’s Theorem). Suppose f is defined in a disk D containing the point (a, b). If the functions fxy and fyx are both continuous on D. then fxy (a, b) = fyx (a, b). We can also define derivatives of order 3 or higher. For instance 2 ∂ ∂ f ∂3f fxyy = (fxy )y = = . ∂y ∂y∂x ∂y 2 ∂x Using Clairaut’s Theorem, one can show fxyy = fyxy = fyyx as long as all partial derivatives of f are continuous. Example 8. Calculate fxxyz if f (x, y, z) = sin(3x + yz). Solution Basic calculation shows fx = 3 cos(3x + yz); fxx = −9 sin(3x + yz); fxxy = −9z cos(3x + yz); fxxyz = −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 ∂2u ∂2u + 2 =0 ∂x2 ∂y is called Laplace’s equation. Solutions to this equation are called harmonic functions. Example 9. Show that the function u(x, y) = ex sin y is a solution to the Laplace’s equation. Solution By a basic calculation we have ux = ex sin y, uy = ex cos y; uxx = ex sin y, uyy = −ey sin y; Therefore we have uxx + uyy = 0. Example 10. Let f be a smooth function. Verify the function u(x, t) = f (x − at) satisfies the wave equation 2 ∂2u 2∂ u = a . ∂t2 ∂x2 4 Solution By a basic calculation we have ut = −af 0 (x − at) ux = f 0 (x − at); utt = a2 f 00 (x − at), uxx = f 00 (x − at); Therefore we have utt = a2 f 00 (x − at) = a2 uxx = 0. 5

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