PhysicsEasy156×since 2002Q8309If Z=A2B3C4Z = {{{A^2}{B^3}} \over {{C^4}}}Z=C4A2B3, then the relative error in Z will be :AΔAA+ΔBB+ΔCC{{\Delta A} \over A} + {{\Delta B} \over B} + {{\Delta C} \over C}AΔA+BΔB+CΔCB2ΔAA+3ΔBB−4ΔCC{{2\Delta A} \over A} + {{3\Delta B} \over B} - {{4\Delta C} \over C}A2ΔA+B3ΔB−C4ΔCC2ΔAA+3ΔBB+4ΔCC{{2\Delta A} \over A} + {{3\Delta B} \over B} + {{4\Delta C} \over C}A2ΔA+B3ΔB+C4ΔCDΔAA+ΔBB−ΔCC{{\Delta A} \over A} + {{\Delta B} \over B} - {{\Delta C} \over C}AΔA+BΔB−CΔCCheck answerSkip