Derivation of the unit change in resistance

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 $${dx \over x}$$.

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 $$R = \rho L^2 / V$$:

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

• Isn't total differential just a linear approximation? A few texts I ran into they all simply take the logarithm then differentiate both sides and somehow these dR, dL etc. just happen to be there for no reason but I find no mention of using total differential or suggesting the equation is an approximation.
– KMC
Feb 1, 2021 at 9:41
• The total differential is a linear approximation, and is only a good one when the changes in the independent variables are small. If I take your original expression for the change in resistance and set $R=V=L=\rho=1$, then let $\mathrm{d}L = 9$, I get $\mathrm{d}R=18$. But you know from the expression for resistance (not the differential) that the real answer is $\Delta R = (\rho / V) \times \Delta(L^2) = 80$. There might be another way of deriving the approximate expression, but the total differential method is valid.
– Tony
Feb 1, 2021 at 10:34

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.