Derivation of the unit change in resistance

Question

The unit change in resistance is given by this equation but I don't understand how it is derived:

$$ {dR \over R} = {d\rho \over \rho} + {2dL \over L} - {dV \over V}. $$

The resistance of a rectangular object is given by

$$ R = \rho {L \over A} = \rho {L^2 \over V}. $$

Taking the logarithm of both sides then differentiating should give

$$ ln(R) = ln(\rho) + 2ln(L) - ln(V) $$ $$ {1 \over R} = {1 \over \rho} + {2 \over L} - {1 \over V} $$

How are the $ dR $, $ dL $, $d\rho$ and $ dV $ come about in the original given equation? $ {d \over dx} ln(x) $ should be $ 1 \over x $, not $ {dx \over x} $.

Answer 1

You need to calculate the total differential of

$$R=\rho \frac{L^2}{V}$$

That is

$$\mathrm{d}R = \frac{\partial R}{\partial \rho}\mathrm{d}\rho + \frac{\partial R}{\partial L} \mathrm{d}L + \frac{\partial R}{\partial V}\mathrm{d} V $$

$$ = \frac{L^2}{V}\mathrm{d}\rho + \frac{2\rho L}{V}\mathrm{d}L - \frac{\rho L^2}{V^2}\mathrm{d}V$$

Then divide by $R = \rho L^2 / V$:

$$\frac{\mathrm{d}R}{R} = \frac{1}{\rho}\mathrm{d}\rho + \frac{2}{L}\mathrm{d}L - \frac{1}{V}\mathrm{d}V$$

Answer 2

You must see R as a function with multivariables, R = R( $\rho$, L, V). Then the change in every direction is given with the total derivative : $$ dR = (\frac{\partial R}{\partial \rho})_{L,V}d\rho + (\frac{\partial R}{\partial L})_{\rho,V}dL +(\frac{\partial R}{\partial V})_{\rho,L}dV $$ Then you will have: $$ dR =d\rho \frac{L^2}{V}+dL \frac{2L\rho}{V}-dV\frac{\rho L^2}{V^2} $$ Dividing both sides with R and using the relation $ R=\rho \frac{ L^2}{V}$ we get the equation.

You must use the total derivative, not the partial derivatives here because you want to find the change of R in every direction, not only one. By including the factors like $ d\rho, dL $ etc you specify the direction in which the change is happening. The importance of factors will be more understandable, if you think R as a vector in 3-dimensional space of $\rho$, L and V.