求函数的偏导数,三题
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(1)s=(u²+v²)/uv=u/v+v/u
∂s/∂u=1/v-v/u²=(u²-v²)/u²v
∂s/∂v=-u/v²+1/u=(v²-u²)/uv²
(2)u=x^(y/z)
∂u/∂x=(y/z)x^(y/z-1)=(y/z)^[(y-z)/z]
∂u/∂y = x^(y/z)lnx*(1/z) = (1/z) * x^(y/z) * lnx,
∂u/∂z = x^(y/z)lnx*(-y/z²) = (-y/z²) * x^(y/z) * lnx
(3)u=(x+y)/(y+z)=x/(y+z) + y/(y+z)
∂u/∂x=1/(y+z)
∂u/∂y=-x/(y+z)² + [y+z-y]/(y+z)²=[-x+z]/(y+z)²
∂u/∂z=-x/(y+z)² - y/(y+z)²=[-x-y]/(y+z)²
∂s/∂u=1/v-v/u²=(u²-v²)/u²v
∂s/∂v=-u/v²+1/u=(v²-u²)/uv²
(2)u=x^(y/z)
∂u/∂x=(y/z)x^(y/z-1)=(y/z)^[(y-z)/z]
∂u/∂y = x^(y/z)lnx*(1/z) = (1/z) * x^(y/z) * lnx,
∂u/∂z = x^(y/z)lnx*(-y/z²) = (-y/z²) * x^(y/z) * lnx
(3)u=(x+y)/(y+z)=x/(y+z) + y/(y+z)
∂u/∂x=1/(y+z)
∂u/∂y=-x/(y+z)² + [y+z-y]/(y+z)²=[-x+z]/(y+z)²
∂u/∂z=-x/(y+z)² - y/(y+z)²=[-x-y]/(y+z)²
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