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A digital thermometer might change by a crisp 0.1Β V0.1\text{ V} per degree at room temperature, but barely nudge by 0.01Β V0.01\text{ V} near freezing. Why does the exact same potential divider circuit become sluggish at different ends of its scale?

Circuit Sensitivity and the Chain Rule

Circuit sensitivity measures how sharply the output voltage responds to a change in a physical quantity like temperature (TT) or illuminance (LL). It is expressed as the derivative dVoutdT\frac{dV_{out}}{dT} or dVoutdL\frac{dV_{out}}{dL} in units like mV/∘C\text{mV}/^\circ\text{C} or mV/lux\text{mV}/\text{lux}.
In 1833, Michael Faraday discovered that silver sulfide decreased in resistance when heated, revealing the first negative temperature coefficient (NTC) behavior. To find how temperature changes translate into voltage changes, we apply the calculus chain rule.
The total circuit sensitivity splits into two distinct conversion stages: the transducer converting physics to resistance, and the divider converting resistance to voltage: dVoutdT=dVoutdRsensΓ—dRsensdT\frac{dV_{out}}{dT} = \frac{dV_{out}}{dR_{sens}} \times \frac{dR_{sens}}{dT}
How exactly does the potential divider circuit dictate the electrical derivative dVoutdRsens\frac{dV_{out}}{dR_{sens}}? Let's differentiate the divider equation directly.

Deriving Electrical Sensitivity

Consider a divider where a variable sensor resistor R1R_1 sits on top and a fixed resistor R2R_2 sits on the bottom. The output voltage across R2R_2 is Vout=Vsβ‹…R2R1+R2V_{out} = V_s \cdot \frac{R_2}{R_1 + R_2}.
Treating R2R_2 and supply voltage VsV_s as constants, we differentiate VoutV_{out} with respect to sensor resistance R1R_1 using the quotient rule: dVoutdR1=ddR1[VsR2(R1+R2)βˆ’1]=βˆ’VsR2(R1+R2)2\frac{dV_{out}}{dR_1} = \frac{d}{dR_1}\left[V_s R_2 (R_1 + R_2)^{-1}\right] = -\frac{V_s R_2}{(R_1 + R_2)^2}
The derivative formula proves that sensitivity is inherently non-linear because R1R_1 appears in the denominator squared. To achieve maximum sensitivity at a specific operating target, you must pick a fixed resistor R2R_2 equal to the sensor resistance R1R_1 at that target point.
What happens when we connect this divider to an analogue-to-digital converter or amplifier that draws current?