1. Partial Derivatives Determine if a limit exists: 1.First test by substituting, : 2.Use two path test to test for non-existence of a limit at a point Two-Path Test for Nonexistence of a limit: If a function f(x,y) has different limits along two different paths as (x,y) approaches (x 0,y 0 ), then lim (x,y) –> (x 0,y 0 ) f(x,y) does not exist. [Note: most often used when (x 0,y 0 ) is the point (0,0). The different paths are defined as y = mx. By evaluating lim (x,y) –> (0, 0) f(x,mx), if m remains in the result, then the limit varies along different paths varies as a function of m and therefore does not exist.] Steps: Set up the generic equation of a line through the point (x 0,y 0 ). If (x 0,y 0 ) is the point (0,0) then the equation is y=mx, otherwise the equation is y = m(x-x 0 ) + y 0. Substitute for y in the original equation. If m doesn’t disappear, then the limit doesn’t exist because the limit varies with the slope of the line through the point (x 0,y 0 ). Example: lim (x,y)  (1,1) = =1/2 xy x 4 + y 2 1*1 1 4 + 1 2 lim (x,y)  (0,0) = lim (x,mx)  (0,0) = lim (x,mx)  (0,0) = xy x 4 + y 2 x*mx x 4 + (mx) 2 mx 2 x 2 (x 2 + m 2 ) 1m1m Partial Derivative with Respect to x: The partial derivative of f(x,y) with respect to x at the point (x 0,y 0 ) is provided the limit exists. Partial Derivative with Respect to y: The partial derivative of f(x,y) with respect to y at the point (x 0,y 0 ) is provided the limit exists. [Note: In many cases,  f/  x   f/  y ]  f f(x 0 +h,y 0 ) – f(x 0,y 0 )  x h 0 h lim (x 0,y 0 ) =f x =  f f(x 0,y 0 +h) – f(x 0,y 0 )  y h 0 h lim (x 0,y 0 ) =f y = Notes: When calculating  f/  x, any y’s in the equation are treated as constants when taking the derivative. Similarly, when calculating  f/  y, any x’s in the equation are treated as constants when taking the derivative. Second order partial derivatives F(x,y,z) = 0 implicitly defines a function z = G(x,y) Implicit Partial Differentiation, e.g. find  z/  x of xz – y ln z = x + y. Treat y’s like constants. f xx = = ( ), f yy = = ( )  2 f   f  2 f   f  x 2  x  x  y 2  y  y f xy = = ( ) = f yx = = ( )  2 f   f  2 f   f  x  y  x  y  y  x  y  x  xz –  y ln z =  x +  y x  z + z  x – y  ln z =  x +  y x  z + z – y  z =  x + 0 dz = 1 – z  x  x  x  x  x  x  x  x  x  x z  x  x  x x – y/z Theorum 5: Chain rule for Functions of Two Independent Variables: If w = f(x,y) has continuous partial derivatives f x and f y and if x = (t) and y = y(t) are differentiable functions of t, then the composite function w = f(x(t), y(t)) is a differentiable function of t and df/dt = f x (x(t), y(t)) * x’(t) + f y (x(t), y(t)) * y’(t), or Theorum 6: Chain Rule for Functions of Three Independent Variables: If w = f(x,y,z) is differentiable and x, y, and z are differentiable functions of t, then w is a differentiable function of t and Theorum 7: Chain Rule for Two Independent Variables and Three Intermediate Variables: Suppose that w = f(x,y,z), x = g(r,s), y = h(r,s), and z = k(r,s). If all four functions are differentiable, then w has partial derivatives with respect to r and s given by the formulas Note the following extensions of Theorum 7: If w = f(x,y), x = g(r,s), and y = h(r,s) then If w = f(x) and x = g(r,s), then Chain Rule dw  f dx  f dy dt  x dt  y dt = + dw  w dx  w dy dt  x dt  y dt = + also written as dw  f dx  f dy  f dz dt  x dt  y dt  z dt = + +  w  w  x  w  y  w  z  r  x  r  y dr  z  r = + +  w  w  x  w  y  w  z  s  x  s  y ds  z  s = + +  w  w  x  w  y  r  x  r  y dr = +  w  w  x  w  y  s  x  s  y ds = +  w dw  x  r dx  r =  w  w  x  s  x  s = Examples: Find the derivative of z = f(x,y) = xy along the curve x = t, y = t 2 at t = 1  z/  x = y  z/  y = x dx/dt = 1 dy/dt = 2t = + = y*1 + x*(2t) substitutu=ing for x and y = t 2 +(t)(2t) = 3t 2 = 3*(1) 2 = 3  z  z dx  z dy  t  x dt  y dt ztzt ztzt t = 1 z t xy zxzx zyzy dx dt dy dt Suppose f(x,y,z) = x + y +z 2 and x = u/v; y = u + ln(v); and z = u. Find  f/  u and  f/  v  f/  x = 1  f/  y = 1  f/  z = 2z  x/  u = 1/v  x/  v = – uv –2  y/  u = 1  y/  v = 1/v  z/  u = 1  z/  v = 0 f(x,y,z) xyzxyz uv fxfx fxfx fyfy fyfy fzfz fzfz xuxu yuyu zuzu xvxv yvyv zvzv  f  f  x  f  y  f  z  u  x  u  y  u  z  u = + +  f  f  x  f  y  f  z  v  x  v  y  v  z  v = + + = 1 * 1/v + 1*1 – 2z * 1 = 1/v + 1 – 2z but z = u = 1/v + 1 + –2u = 1 * (–uv –2 ) + 1*1/v – 2z * 0 = (–uv –2 ) + 1/v