证明存在s属于1到2,fs的导数-fs=f2-2f1答:f(x)在[1,3]上二阶可导,因此f(x)与f’(x)在[1,3]上连续 在[1,2]上对f(x)运用拉格朗日中值定理,存在一点ξ₁∈(1,2),使得 f(2)- f(1)=(2-1)*f’(ξ₁)=f’(ξ₁)∵f(2)>f(1),∴f’(ξ₁)>0 由于f(2)>∫{2,3}f(x)dx,利用积分中值...
为什么二阶偏导数连续 ,混合偏导就相等啊??答:F(x,y)=x^3y^3sin(1/(xy)),xy≠0. F(x,y)=0,xy=0. 1.xy=0,显然有 Fx'(x,y)=Fy'(x,y)=0. 2.xy≠0, Fx'(x,y)=3x^2y^3sin(1/(xy))-xy^2cos(1/(xy)), Fy'(x,y)=3x^3y^2sin(1/(xy))-x^2ycos(1/(xy)). 3. xy=0,显然有 Fxy''(x,y)=Fyx''(...