设f（x,y)=x^2-xy+2y^2,求f(1+Δx，y)和f（x，2-Δy）

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摘要 df(x, y) = {[xy(x^2 - y^2)]'(x^2 + y^2) - xy(x^2 - y^2)(x^2 + y^2)'} / (x^2 + y^2)^2

= {[(xy)'(x^2 - y^2) + (xy)(x^2 - y^2)'](x^2 + y^2) - xy(x^2 - y^2)(2xdx + 2ydy)} / (x^2 + y^2)^2

= {[(ydx + xdy)(x^2 - y^2) + xy(2xdx - 2ydy)](x^2 + y^2) - 2xy(x^2 - y^2)(xdx + ydy)} / (x^2 + y^2)^2

= [y(x^4 - 4x^2y^2 - y^4) / (x^2 + y^2)^2]dx - [x(x^4 - 5y^4) / (x^2 + y^2)^2]dy

咨询记录 · 回答于2024-01-04

设f（x,y)=x^2-xy+2y^2,求f(1+Δx，y)和f（x，2-Δy）

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您好很高兴为您解答设函数f(x,y)={xy(x^2-y^2)/(x^2+y^2) ,当(x,y) ≠(0,0);0,当（x,y）=(0,0).求f”yx(0,0),f”xy(0,0).

df(x, y) = [xy(x^2 - y^2)]'(x^2 + y^2) - xy(x^2 - y^2)(x^2 + y^2)' df(x, y) = [(xy)'(x^2 - y^2) + (xy)(x^2 - y^2)'](x^2 + y^2) - xy(x^2 - y^2)(2xdx + 2ydy) df(x, y) = [(ydx + xdy)(x^2 - y^2) + xy(2xdx - 2ydy)](x^2 + y^2) - 2xy(x^2 - y^2)(xdx + ydy) df(x, y) = [y(x^4 - 4x^2y^2 - y^4)/(x^2 + y^2)^2]dx - [x(x^4 - 5y^4)/(x^2 + y^2)^2]dy

f''x,y(0,0)=-1f''xy,x(0,0)=1

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第一题啊