高数求二阶偏导如图所示
1个回答
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解:
z=f(x+y,xy)
∂z/∂x
=f'1·1+f'2·y
=f'1+yf'2
∂²z/∂x²
=(f''11·1+f''12·y)+y(f''21·1+f''22·y)
=f''11+yf''12+yf''21+y²f''22
=f''11+2yf''12+y²f''22
∂²z/∂x∂y
=f''11+f''12·x+y(f''21+f''22·x)
=f''11+(x+y)f''12+xyf''22
∂z/∂y
=f'1+xf'2
∂²z/∂y²
=f''11+xf''12+x(f''21+xf''22)
=f''11+2xf''12+x²f''22
z=f(x+y,xy)
∂z/∂x
=f'1·1+f'2·y
=f'1+yf'2
∂²z/∂x²
=(f''11·1+f''12·y)+y(f''21·1+f''22·y)
=f''11+yf''12+yf''21+y²f''22
=f''11+2yf''12+y²f''22
∂²z/∂x∂y
=f''11+f''12·x+y(f''21+f''22·x)
=f''11+(x+y)f''12+xyf''22
∂z/∂y
=f'1+xf'2
∂²z/∂y²
=f''11+xf''12+x(f''21+xf''22)
=f''11+2xf''12+x²f''22
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