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If $Z = \frac{A^{2} B^{3}}{C^{4}}$ then the relative error in z will be then the relative error in z will be

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$\frac{Δ A}{A} + \frac{Δ B}{B} + \frac{Δ C}{C}$

$2 \frac{Δ A}{A} + 3 \frac{Δ B}{B} - 4 \frac{Δ C}{C}$

$2 \frac{Δ A}{A} + 3 \frac{Δ B}{B} + 4 \frac{Δ C}{C}$

$\frac{Δ A}{A} + \frac{Δ B}{B} - \frac{Δ C}{C}$

answer is C.

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Detailed Solution

$Z = \frac{A^{2} B^{3}}{C^{4}}$

For the total error in Z.

$\frac{dZ}{Z} = \frac{2 dA}{A} + \frac{3 dB}{B} + \frac{4 dC}{C}$

$\Rightarrow \frac{ΔZ}{Z} = \frac{2 ΔA}{A} + \frac{3 ΔB}{B} + \frac{4 ΔC}{C}$

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