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Partial Derivatives Chain Rule Determine if a limit exists: 1. 2. xy 1*1 = 4 =1/2 x4 + y2 1 + 12 Use two path test to test for non-existence of a limit at a point First test by substituting, : lim (x,y) (1,1) 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: lim (x,y) (0,0) xy = lim x4 + y2 (x,mx)(0,0) x*mx = lim x4 + (mx)2 (x,mx)(0,0) mx2 = x2(x2 + m2) 1 m 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 dw df dx df dy dw dw dx dw dy also written as dt = dx dt + dy dt dt = dx dt + dy dt 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 dw df dx df dy df dz dt = dx dt + dy dt +dz dt 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 dw dw dx dw dy dw dz dw dw dx dw dy dw dz dr = dx dr + dy dr + dz dr ds = dx ds + dy ds + dz ds Note the following extensions of Theorum 7: If w = f(x,y), x = g(r,s), and y = h(r,s) then dw dw dx dw dy dw dw dx dw dy dr = dx dr + dy dr ds = dx ds + dy ds If w = f(x) and x = g(r,s), then dw dw dx dw dw dx dr = dx dr ds = dx ds Examples: Partial Derivative with Respect to x: The partial derivative of f(x,y) with df f(x0+h,y0) – f(x0,y0) lim respect to x at the point (x0,y0) is fx = = dx (x0,y0) h Find the derivative of z = f(x,y) = xy along the curve x = t, y = t2 at t = 1 dz dx x 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 df f(x0,y0+h) – f(x0,y0) fy = dy (x0,y0) = h 0 dz dy y dx dt lim h dz/dx = y z h 0 dz/dy = x dx/dt = 1 dy/dt = 2t dz dz dx dz dy = + = y*1 + x*(2t) dt dx dt dy dt dz substitutu=ing for x and y = t2 +(t)(2t) = 3t2 dt dz = 3*(1)2 = 3 dt dy dt provided the limit exists. [Note: In many cases, df/dx df/dy ] t t=1 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 d2f d df d2f d df d2f d fxx = = , fyy = = fxy = = dx2 dx dx dy2 dy dy dx dy dx •F(x,y,z) = 0 implicitly defines a function z = G(x,y) ( ) ( ) ( dfdy)= f yx = d2f d df = dy dx dy dx ( ) •Implicit Partial Differentiation, e.g. find dz/dx of xz – y ln z = x + y. Treat y’s like constants. dxz – dy ln z = dx + dy dx dx dx dx xdz + zdx – ydln z = dx + dy dx dx dx dx dx xdz + z – y dz = dx + 0 dx z dx dx dz = 1 – z dx x – y/z Suppose f(x,y,z) = x + y f(x,y,z) df dx x df dy dy dx du du u df dz y +z2 and x = u/v; y = u + ln(v); and z = u. Find df/du and df/dv f(x,y,z) df/dx = 1 df/dy = 1 df dx z df dy x dz du dx dv df df dx df dy df dz = + + du dx du dy du dz du dy dv v df dz y df/dz = 2z z dz dv dx/du = 1/v dx/dv = – uv –2 dy/du = 1 dy/dv = 1/v dz/du = 1 dz/dv = 0 df df dx df dy df dz = + + dv dx dv dy dv dz dv = 1 * 1/v + 1*1 – 2z * 1 = 1 * (–uv –2) + 1*1/v – 2z * 0 = 1/v + 1 – 2z but z = u = (–uv –2) + 1/v = 1/v + 1 + –2u