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∂;z/∂;x則把y看成常數x*1/z=ln(z/y)所以1/z∂;x+x*(-1/z²;)∂;z=1/(z/y)*(1/y)∂;z1/z∂;x-x/z²;∂;z=(1/z)∂;z所以∂;z/∂;x=(1/z)/(1/z+x/z²;)=z /(x+z)同…
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