Hi! Does anyone know thy the percentage change in of two numbers multiplied together is roughly equal to the percentage change in the first number plus the percentage change in the second number? How do you prove that?
Hi! Does anyone know thy the percentage change in of two numbers multiplied together is roughly equal to the percentage change in the first number plus the percentage change in the second number? How do you prove that?
First it makes no sense to talk about "percentage change" without saying "percentage of what". I suspect that you are referring to what I would call the "relative change", the change in each quantity divided by itself. There is an old "engineer's rule of thumb" that "If you are adding two measured quantities, the 'errors in the measurement' add and if you multiply two measured quantities the relative errors add".
But that cannot be proved because it is not exactly true- it is approximately true. That can be shown using the "product rule". Suppose two quantities, A and B, increase by "infinitesimal" amounts dA and dB, respectively. Then, for any parameter, t, the product rule, applied to AB, $\displaystyle \frac{dAB}{dt}= \frac{dA}{dt}B+ A\frac{dB}{dt}$. Ignoring the "dt" gives the differential form $\displaystyle d(AB)= (dA)B+ A(dB)$. Finally, dividing both sides by AB gives $\displaystyle \frac{d(AB)}{AB}= \frac{dA}{A}+ \frac{dB}{B}$.
But that is "approximate" because the actual changes in AB, A, and B are only approximately equal to d(AB), dA, and dB respectively.
Suppose that the two numbers $\displaystyle A$ and $\displaystyle B$ are subject to small changes $\displaystyle \delta A$ and $\displaystyle \delta B$ respectively, then the resulting small change in their product $\displaystyle AB,\: \delta(AB),$ will be given by
$\displaystyle \delta(AB)=(A+\delta A)(B+\delta B)-AB =B\delta A+A\delta B + \delta A. \delta B.$
Dividing throughout by $\displaystyle AB,$
$\displaystyle \frac{\delta(AB)}{AB}=\frac{\delta A}{A}+\frac{\delta B}{B}+\frac{\delta A.\delta B}{AB},$
and on the assumption that the third term on the RHS is negligible compared with the first two terms on the RHS,
$\displaystyle \frac{\delta(AB)}{AB}\approx \frac{\delta A}{A}+\frac{\delta B}{B}.$
(Think numbers, if $\displaystyle \delta A/A $ and $\displaystyle \delta B/B$ are of the order $\displaystyle 10^{-3}$ say, then their product will be of the order $\displaystyle 10^{-6}.)$
Multiplying throughout by 100 produces an approximate percentage relationship.