2022-01-06 11:50发布
贵州大学数学研究生怎么样
贵州大学数学研究生怎么样 对方程z = f(y/x,x+2y)的两端求微分,得版dz = f1*[(xdy-ydx)/x²]+f2*(dx+2dy)= [-(y/x²)f1+f2]dx+[(1/x)f1+2*f2]dy,得到Dz/Dx = -(y/x²)f1+f2,Dz/Dy = (1/x)f1+2*f2,于是权D²z/DxDy = (D/Dx)(Dz/Dy)= (D/Dx)[(1/x)f1+2*f2]= [(-1/x²)*f1+(1/x)*[-(y/x²)f11+f12]+2*[(1/x)f21+2*f22]
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贵州大学数学研究生怎么样 对方程z = f(y/x,x+2y)的两端求微分,得版dz = f1*[(xdy-ydx)/x²]+f2*(dx+2dy)= [-(y/x²)f1+f2]dx+[(1/x)f1+2*f2]dy,得到Dz/Dx = -(y/x²)f1+f2,Dz/Dy = (1/x)f1+2*f2,于是权D²z/DxDy = (D/Dx)(Dz/Dy)= (D/Dx)[(1/x)f1+2*f2]= [(-1/x²)*f1+(1/x)*[-(y/x²)f11+f12]+2*[(1/x)f21+2*f22]
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