Section 10.3: Partial Derivatives It is often useful to know how a function of several variables changes with respect to one of its variables. Definition: Suppose that f (x, y) is a function of two variables. The partial derivatives of f with respect to x and y are defined by f (x + h, y) − f (x, y) , h→0 h f (x, y + h) − f (x, y) fy (x, y) = lim , h→0 h fx (x, y) = lim provided these limits exist. Example: Find the partial derivatives of f (x, y) = x2 − 4xy + y 2 using the limit definition. 1 How to Find the Partial Derivatives of f (x, y): 1. To find fx (x, y), treat y as a constant and differentiate f with respect to x. 2. To find fy (x, y), treat x as a constant and differentiate f with respect to y. Notation: If z = f (x, y), the partial derivatives of f with respect to x and y are denoted by ∂z ∂f = = zx = Dx f, ∂x ∂x ∂f ∂z fy = = = zy = Dy f. ∂y ∂y fx = Example: Find the partial derivatives fx (x, y) and fy (x, y) for each function. (a) f (x, y) = x3 + x2 y + xy 2 − 3y 3 (b) f (x, y) = ln(3x2 − xy) (c) f (x, y) = xexy + sin(3x2 + y 3 ) 2 Example: Find zx and zy if z is defined implicitly by the equation x2 z + yz 3 = x + y 2 . Geometric Interpretation of Partial Derivatives: The equation z = f (x, y) defines a surface S. If f (x0 , y0 ) = z0 , then the point (x0 , y0 , z0 ) lies on S. The vertical plane x = x0 intersects S in a curve C1 parallel to the yz-plane. An equation for this curve is z = f (x0 , y). The slope of the tangent line to this curve at (x0 , y0 , z0 ) is fy (x0 , y0 ). Similarly, the vertical plane y = y0 intersects S in a curve C2 parallel to the xz-plane. An equation for C2 is z = f (x, y0 ). The slope of the tangent line to C2 at (x0 , y0 , z0 ) is fx (x0 , y0 ). Thus, partial derivatives can be interpreted as rates of change in the x and y directions, respectively. 3 Note: The definitions of partial derivatives can be extended to functions of three or more variables. Example: Find the partial derivatives fx , fy , and fz for each function. (a) f (x, y, z) = x5 + x4 y 4 z 3 + yz 2 (b) f (x, y, z) = y tan(x2 + z) (c) f (x, y, z) = ln(x2 y + y 2 z + 2xz 3 ) 4 Definition: Suppose that f (x, y) is a function of two variables. The second-order partial derivatives of f are ∂ ∂f ∂ 2f fxx = (fx )x = = , ∂x ∂x ∂x2 ∂ 2f ∂ ∂f = fyy = (fy )y = , ∂y ∂y ∂y 2 ∂ ∂f ∂ 2f fxy = (fx )y = = , ∂y ∂x ∂y∂x ∂ ∂f ∂ 2f fyx = (fy )x = = . ∂x ∂y ∂x∂y The second-order partial derivatives fxy and fyx are called mixed partial derivatives. Example: Find all second-order partial derivatives of f (x, y) = x2 y + sin(x + y). Example: Find the mixed partial derivatives of f (x, y) = ln(x2 + 3xy). 5 Theorem: (The Mixed-Partials Theorem) If f (x, y) and its partial derivatives fx , fy , fxy , and fyx are continuous on an open disk centered at (x0 , y0 ), then fxy (x0 , y0 ) = fyx (x0 , y0 ). Example: Suppose that f (x, y) = tan−1 (x2 y). Show that fxy (x, y) = fyx (x, y) for all (x, y) in the domain of f . 6