Statement I: If $x=\frac{a^{m} \cdot b^{n}}{c^{p} \cdot d^{q}}$, then the percentage error in the me

Question

Statement I: If $x=\frac{a^{m} \cdot b^{n}}{c^{p} \cdot d^{q}}$, then the percentage error in the measurement of $x$,\ $\frac{\Delta x}{x}=m\left(\frac{\Delta a}{a}ight)+n\left(\frac{\Delta b}{b}ight)+p\left(\frac{\Delta c}{c}ight)+q\left(\frac{\Delta d}{d}ight)$\ Statement II: The above is true for all values of $\Delta a, \Delta b, \Delta c$ and $\Delta d$.\

Professor PC Thomas Classes · Answer

(b) : Statement I is correct, Statement II is incorrect. $\log x=m \log a+n \log b-p \log c-q \log d$ $ \frac{d x}{x}=m\left(\frac{d a}{a}ight)+n\left(\frac{d b}{b}ight)-p\left(\frac{d c}{c}ight)-q\left(\frac{d d}{d}ight) $ \  We are doing a differential approximation here, by putting $d x=\Delta x, d a=\Delta a, d b=\Delta b, d c=\Delta c$ and $d d=\Delta d$. So $\Delta a, \Delta b$, $\Delta c, \Delta d$ must be small. Also, the errors could be +ve or -ve , so for maximum possible error we put everywhere as +ve . So, \ $ \frac{\Delta x}{x}=m\left(\frac{\Delta a}{a}ight)+n\left(\frac{\Delta b}{b}ight)+p\left(\frac{\Delta c}{c}ight)+q\left(\frac{\Delta d}{d}ight) $