If Q = frac{X^n}{Y^m} and Delta X is the abso | vedclass

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(B) Given the expression $Q = \frac{X^n}{Y^m}$.Taking the natural logarithm on both sides: $\ln Q = n \ln X - m \ln Y$.Differentiating both sides,we g

Vedclass · Accepted Answer

$\Delta Q = \pm \left( n\frac{\Delta X}{X} + m\frac{\Delta Y}{Y} \right) Q$

Vedclass · Answer

(B) Given the expression $Q = \frac{X^n}{Y^m}$.Taking the natural logarithm on both sides: $\ln Q = n \ln X - m \ln Y$.Differentiating both sides,we get the expression for relative error: $\frac{dQ}{Q} = n \frac{dX}{X} - m \frac{dY}{Y}$.For maximum absolute error,we consider the sum of the magnitudes of the relative errors: $\frac{\Delta Q}{Q} = \pm \left( n \frac{\Delta X}{X} + m \frac{\Delta Y}{Y} \right)$.Therefore,the absolute error $\Delta Q$ is given by: $\Delta Q = \pm \left( n \frac{\Delta X}{X} + m \frac{\Delta Y}{Y} \right) Q$.

If $Q = \frac{X^n}{Y^m}$ and $\Delta X$ is the absolute error in the measurement of $X,$ and $\Delta Y$ is the absolute error in the measurement of $Y,$ then the absolute error $\Delta Q$ in $Q$ is:

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