Determine the secondorder Taylor formula for the given funct

Solution

f(x,y) = e^(x^2 - 2x + 1) * cos y ?f / ?x = (2x-2) e^(x^2 - 2x + 1) * cos y ?f / ?y = - e^(x^2 - 2x + 1) * sin y ?²f / ?x² = (2x-2)^2 e^(x^2 - 2x + 1) * cos y + 2e^(x^2 - 2x + 1) * cos y = (4x^2 - 8x + 6) e^(x^2 - 2x + 1) * cos y [this is the one different from the one you have shown, you forgot the product rule in x.] ?²f / ?x?y = -(2x-2) e^(x^2 - 2x + 1) * sin y ?²f / ?y² = - e^(x^2 - 2x + 1) * cos y the above are the correct partial derivatives ... f(1,0) = 1 fx(1,0) = 0 fy(1,0) = 0 fxx(1,0) = 2 fxy(1,0) = 0 fyy(1,0) = -1 Now, f(x,y) ˜ f(1,0) + fx(1,0) (x-1) + fy(1,0) (y - 0) + 1/2 [fxx(1,0) (x-1)^2 + 2 fxy(1,0) (x-1)(y-0) + fyy(1,0) (y-0)^2] + ... = 1 + 0 + 0 + 1/2 [ 2 (x-1)^2 - y^2 ]