How do you find the df/du AND df/dv of sin(xy)+xsiny (when x=(u^2+v^2), y=(uv))?

2012-12-06 3:20 pm
I've been puzzled over this problem for two weeks. Also, u=0 and v=1.

回答 (1)

2012-12-06 4:11 pm
✔ 最佳答案
F=sin(xy)+xsin(y)=>
dF/du=Fx(Xu)+Fy(Yu)=>
dF/du=[ycos(xy)+sin(y)](2u)+
[xcos(xy)+xcos(y)]v=>

dF/du=2u{uvcos[(u^2+v^2)uv]+
sin(uv)}+v{(u^2+v^2)cos[(u^2+v^2)uv]+
(u^2+v^2)cos(uv)}
When u=0,v=1
dF/du=cos(0)+cos(0)=2

Similarly, you can find dF/dv yourself.


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