加減證明
無法想出一個
從公式開始：因變數等於常數乘以自變數 y = x ∗ z {\displaystyle y=x*z} 那麼 δ y = ( δ x ∂ ( x ∗ z ) ∂ x ) 2 + ( δ z ∂ ( x ∗ z ) ∂ z ) 2 = ( z δ x ) 2 + ( x δ z ) 2 {\displaystyle \delta _{y}={\sqrt {\left(\delta _{x}{\frac {\partial {\left(x*z\right)}}{\partial x}}\right)^{2}+\left(\delta _{z}{\frac {\partial (x*z)}{\partial z}}\right)^{2}}}={\sqrt {(z\delta _{x})^{2}+(x\delta _{z})^{2}}}}
y = x / z {\displaystyle y=x/z} 然後 δ y = ( δ x ∂ ( x z ) ∂ x ) 2 + ( δ z ∂ ( x z ) ∂ z ) 2 = ( δ x z ) 2 + ( − x δ z z 2 ) 2 = ( δ x z ) 2 + ( x δ z z 2 ) 2 {\displaystyle \delta _{y}={\sqrt {\left(\delta _{x}{\frac {\partial {\left({\frac {x}{z}}\right)}}{\partial x}}\right)^{2}+\left(\delta _{z}{\frac {\partial ({\frac {x}{z}})}{\partial z}}\right)^{2}}}={\sqrt {\left({\frac {\delta _{x}}{z}}\right)^{2}+\left({\frac {-x\delta _{z}}{z^{2}}}\right)^{2}}}={\sqrt {\left({\frac {\delta _{x}}{z}}\right)^{2}+\left({\frac {x\delta _{z}}{z^{2}}}\right)^{2}}}}
