Question
Find the relative error in Z if $\text{Z}=\frac{\text{A}^4\text{B}^{\frac{1}{3}}}{\text{CD}^{\frac{3}{2}}}$

Answer

Here, $​​\text{Z}=\frac{\text{A}^4\text{B}^{\frac{1}{3}}}{\text{CD}^{\frac{3}{2}}}$
Relative error, $=\frac{\Delta\text{Z}}{\text{Z}}$
$=\pm\Big[4\Big(\frac{\Delta\text{A}}{\text{A}}\Big)+\frac{1}{3}\Big(\frac{\Delta\text{B}}{\text{B}}\Big)+\Big(\frac{\Delta\text{C}}{\text{C}}\Big)+\frac{3}{2}\Big(\frac{\Delta\text{D}}{\text{D}}\Big)\Big]$

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