Partial Derivatives Chain Rule Determine if a limit exists: First test by substituting, : 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 (x0,y0), then lim (x,y) –> (x0,y0) f(x,y) does not exist. [Note: most often used when (x0,y0) 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 (x0,y0). If (x0,y0) is the point (0,0) then the equation is y=mx, otherwise the equation is y = m(x-x0) + y0. 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 (x0,y0). Example: Theorum 5: Chain rule for Functions of Two Independent Variables: If w = f(x,y) has continuous partial derivatives fx and fy 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 = fx(x(t), y(t)) * x’(t) + fy(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 lim (x,y)  (1,1) = =1/2 xy x4 + y2 1*1 14 + 12 dw df dx df dy dt dx dt dy dt = + dw dw dx dw dy dt dx dt dy dt = + also written as dw df dx df dy df dz dt dx dt dy dt dz dt = + + dw dw dx dw dy dw dz dr dx dr dy dr dz dr = + + dw dw dx dw dy dw dz ds dx ds dy ds dz ds = + + dw dw dx dw dy dr dx dr dy dr = + dw dw dx dw dy ds dx ds dy ds = + lim (x,y)  (0,0) = lim (x,mx)(0,0) = lim (x,mx)(0,0) = xy x4 + y2 x*mx x4 + (mx)2 mx2 x2(x2 + m2) 1m dw dw dx dr dx dr = dw dw dx ds dx ds = Examples: Find the derivative of z = f(x,y) = xy along the curve x = t, y = t2 at t = 1 dz/dx = y dz/dy = x dx/dt = 1 dy/dt = 2t = + = y*1 + x*(2t) substitutu=ing for x and y = t2 +(t)(2t) = 3t2 = 3*(1)2 = 3 Partial Derivative with Respect to x: The partial derivative of f(x,y) with respect to x at the point (x0,y0) 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 (x0,y0) is provided the limit exists. [Note: In many cases, df/dx  df/dy ] df f(x0+h,y0) – f(x0,y0) dx h 0 h lim (x0,y0) = fx = z dz dx dz dy x y df f(x0,y0+h) – f(x0,y0) dy h 0 h lim (x0,y0) = fy = dx dt dy dt dz dz dx dz dy dt dx dt dy dt dzdt t dzdt Notes: When calculating df/dx, any y’s in the equation are treated as constants when taking the derivative. Similarly, when calculating df/dy, 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 dz/dx of xz – y ln z = x + y. Treat y’s like constants. fxx = = ( ), fyy = = ( ) d2f d df d2f d df dx2 dx dx dy2 dy dy fxy = = ( )= fyx = = ( ) d2f d df d2f d df dx dy dx dy dy dx dy dx dxz – dy ln z = dx + dy xdz + zdx – ydln z = dx + dy xdz + z – y dz = dx + 0 dz = 1 – z dx dx dx dx dx dx dx dx dx dx z dx dx dx x – y/z t = 1 Suppose f(x,y,z) = x + y +z2 and x = u/v; y = u + ln(v); and z = u. Find df/du and df/dv df/dx = 1 df/dy = 1 df/dz = 2z dx/du = 1/v dx/dv = – uv –2 dy/du = 1 dy/dv = 1/v dz/du = 1 dz/dv = 0 f(x,y,z) f(x,y,z) df dx df dz df dx df dz df dy df dy x y z x y z dy du dy dv dx du dz du dx dv dz dv u v df df dx df dy df dz du dx du dy du dz du = + + df df dx df dy df dz dv dx dv dy dv dz dv = + + = 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