设函数u(x，y)=arctan(y/x)，则du∣_(3,4)=____.

设函数u(x，y)=arctan(y/x)，则du∣_(3,4)=____.

-(4/25)dx+(3/25)dy. 解析:∂u/∂x=1/[1+(y/x)²]•[-(y/ x²)]= -[y/(x²+y²)] ∂u/∂y=1/[1+(y/x)²]•(1/x)=x/(x²+y²) 所以∂u/∂x∣_(3,4)=-[4/(3²+4²)]=-(4/25), ∂u/∂y∣_(3,4)=3/25， 故 du∣_(3,4)=∂u/∂x∣_(3,4)dx+∂u/∂y∣_(3,4)dy=-(4/25)dx+(3/25)dy．