Local Sensitivity Derivatives Around a Reference Case

Why Local Derivatives Matter

A decision model may combine pharmacometric parameters, predicted clinical outcomes and a decision quantity such as a benefit score. A local sensitivity derivative answers a focused question: if one input changes slightly from a specified reference case, how quickly does the decision outcome change? It is a rate of change at that case, not a summary of every possible input value or scenario.

This makes derivatives useful for interpreting model behavior near a chosen set of assumptions. They can show which direction of change would raise or lower an outcome, and give a first approximation to the effect of a small parameter perturbation. Their scope is deliberately local: a derivative at one point need not represent behavior far from that point.

Definition: Let $f(\boldsymbol{\theta})$ be a scalar decision outcome, where $\boldsymbol{\theta}=(\theta_1,\ldots,\theta_k)$ is a vector of model inputs or parameters. If $f$ is differentiable at a reference case $\boldsymbol{\theta}_0$, its local sensitivity to $\theta_j$ is the partial derivative $$ \left.\frac{\partial f}{\partial \theta_j}\right|_{\boldsymbol{\theta}_0} = \lim_{h\to 0} \frac{f(\theta_{01},\ldots,\theta_{0j}+h,\ldots,\theta_{0k})-f(\boldsymbol{\theta}_0)}{h}. $$ All other inputs are held fixed in this partial derivative.

The derivative’s units are the units of $f$ divided by the units of $\theta_j$. Its sign describes the local direction of change: a positive derivative means that a small increase in $\theta_j$, with other inputs fixed, locally increases $f$; a negative derivative means it locally decreases $f$. The magnitude describes the rate of change in those units. It is not, on its own, a measure of the parameter’s uncertainty or of its overall importance across a range of plausible values.

First-Order Approximation Around a Reference Case

For a small change $\Delta\boldsymbol{\theta}$, the gradient collects the partial derivatives for all inputs. The first-order Taylor approximation estimates the resulting change in the outcome:

f(\boldsymbol{\theta}_0+\Delta\boldsymbol{\theta}) \approx f(\boldsymbol{\theta}_0) + \nabla f(\boldsymbol{\theta}_0)^\mathsf{T}\Delta\boldsymbol{\theta}.

If only $\theta_j$ changes, this reduces to $f(\boldsymbol{\theta}_0+\Delta\theta_j\mathbf{e}_j)\approx f(\boldsymbol{\theta}_0)+(\partial f/\partial\theta_j)\Delta\theta_j$, where $\mathbf{e}_j$ selects that input. The approximation is most useful when the outcome is differentiable near the reference case and the perturbation is small enough that curvature does not materially alter the rate of change.

Justification: If $f$ has continuous second derivatives along the line segment from $\boldsymbol{\theta}_0$ to $\boldsymbol{\theta}_0+\Delta\boldsymbol{\theta}$, Taylor’s theorem gives a remainder of the form $$ f(\boldsymbol{\theta}_0+\Delta\boldsymbol{\theta}) = f(\boldsymbol{\theta}_0) +\nabla f(\boldsymbol{\theta}_0)^\mathsf{T}\Delta\boldsymbol{\theta} +\frac{1}{2}\Delta\boldsymbol{\theta}^\mathsf{T} H(\boldsymbol{\theta}_0+t\Delta\boldsymbol{\theta}) \Delta\boldsymbol{\theta} $$ for some $t$ between 0 and 1, where $H$ is the Hessian matrix of second derivatives. If the Hessian’s operator norm is at most $M$ along that segment, the absolute remainder is at most $\frac{1}{2}M\|\Delta\boldsymbol{\theta}\|^2$. Thus, a small perturbation and limited curvature support the linear approximation; a large perturbation or substantial curvature may not.

The derivative is defined by changing one input while holding the others fixed. If the actual change of interest must follow a relationship among inputs, use a directional derivative instead: for a specified direction $\mathbf{v}$, the local rate along that direction is $\nabla f(\boldsymbol{\theta}_0)^\mathsf{T}\mathbf{v}$. This distinction matters when parameters are constrained or when a scenario changes several assumptions together.

Worked Example: Sensitivity of an Emax Response to EC50

Hypothetical data. At a fixed dose of 100 mg, suppose the predicted response is $R=E_{\max}d/(EC_{50}+d)$, where $E_{\max}=80$ response percentage points and $EC_{50}=50$ mg. Find the local derivative with respect to $EC_{50}$, then estimate and check the effect of increasing $EC_{50}$ by 5 mg.

At the reference case, $R=80(100)/(50+100)=8000/150=53.33$ response percentage points. Differentiating with respect to $EC_{50}$, with $E_{\max}$ and $d$ fixed, gives

\frac{\partial R}{\partial EC_{50}} = -\frac{E_{\max}d}{(EC_{50}+d)^2} = -\frac{80(100)}{(50+100)^2} = -0.3556\ \text{response percentage points per mg}.

The local approximation for a 5 mg increase is $\Delta R\approx(-0.3556)(5)=-1.778$ response percentage points, giving an approximate response of $53.33-1.778=51.56$ percentage points. Direct evaluation at $EC_{50}=55$ mg gives $R=8000/155=51.61$ percentage points, a change of $51.61-53.33=-1.72$ percentage points. The approximation is close for this perturbation, but it is not exact.

Derivatives Through a Pharmacometric Model

A decision outcome often depends on a parameter through an intermediate prediction. In that case, the chain rule connects the sensitivity of the outcome to the sensitivity of the prediction. For example, if exposure $C$ depends on clearance $CL$, and the response depends on $C$, then $\partial R/\partial CL=(\partial R/\partial C)(\partial C/\partial CL)$, provided these derivatives exist at the reference case.

Worked Example: Propagating Clearance Sensitivity to Response

Hypothetical model. At a fixed dose of 100 dose units, let the exposure proxy be $C=d/CL$, with $CL$ measured in L/h and $C$ in exposure units. Let response be $R=E_{\max}C/(EC_{50}+C)$, with $E_{\max}=60$ response percentage points and $EC_{50}=5$ exposure units. At the reference $CL=10$ L/h, calculate the local response derivative with respect to clearance and estimate the effect of increasing clearance to 11 L/h.

First, $C=100/10=10$ exposure units. The response is $R=60(10)/(5+10)=40.00$ response percentage points. The component derivatives are

\frac{\partial R}{\partial C} = \frac{E_{\max}EC_{50}}{(EC_{50}+C)^2} = \frac{60(5)}{(5+10)^2} = 1.333\ \frac{\text{response percentage points}}{\text{exposure unit}}, \qquad \frac{\partial C}{\partial CL} = -\frac{d}{CL^2} = -\frac{100}{10^2} = -1.000\ \frac{\text{exposure units}}{\text{L/h}}.

By the chain rule, $\partial R/\partial CL=(1.333)(-1.000)=-1.333$ response percentage points per L/h. For a 1 L/h increase, the linear approximation predicts a response change of $-1.333$ points and a response of $38.67$ points. Direct calculation at $CL=11$ L/h gives $C=100/11=9.091$ exposure units and $R=60(9.091)/(5+9.091)=38.71$ points. The direct change is about $-1.29$ points. The negative sign is consistent with this model: higher clearance lowers the exposure proxy, which lowers the response.

Units, Scaling, and Decision Outcomes

A derivative’s magnitude depends on the units used for the input. If a parameter is expressed in mg rather than g, its numerical derivative changes by a factor of 1,000 even though the underlying model behavior does not. Always report units and the reference value. When comparing parameters on different scales, a dimensionless local sensitivity can help, if the reference outcome is nonzero:

S_j = \frac{\theta_{0j}}{f(\boldsymbol{\theta}_0)} \left.\frac{\partial f}{\partial\theta_j}\right|_{\boldsymbol{\theta}_0}.

This quantity approximates the fractional change in $f$ for a fractional change in $\theta_j$ near the reference case. For example, $S_j=-0.4$ indicates that a 1% increase in $\theta_j$ corresponds locally to an approximately 0.4% decrease in $f$, with other inputs fixed. This interpretation requires nonzero $f(\boldsymbol{\theta}_0)$; it is also unhelpful when $\theta_{0j}=0$. In those cases, report the ordinary derivative or choose explicit, meaningful input and outcome scales instead.

For a decision outcome formed from multiple predicted quantities, differentiate the stated outcome rather than an isolated model prediction and assume it represents the whole decision. For example, a weighted benefit-minus-harm score has sensitivities determined by both the model predictions and their weights. A derivative of one component alone cannot establish the sensitivity of the combined score.

Worked Example: Sensitivity of a Benefit-minus-Harm Score

Hypothetical decision rule. Suppose a score is $U=B-40p$, where $B$ is predicted benefit in benefit points and $p$ is the probability of a specified harm. The factor 40 represents 40 benefit points per unit probability of harm. At the reference case, $B=12$ points and $p=0.08$. Find the score and its local sensitivity to harm probability, then estimate the effect of increasing $p$ by 0.01.

The reference score is $U=12-40(0.08)=8.80$ benefit points. Since $B$ is fixed for this partial derivative, $\partial U/\partial p=-40$ benefit points per unit probability. A 0.01 increase in $p$ therefore gives the exact change $\Delta U=(-40)(0.01)=-0.40$ points, so the new score is $8.40$ points. Here the linear calculation is exact because $U$ is linear in $p$. The derivative alone does not indicate how plausible that probability change is; it describes the score’s response to the change.

When a Local Approximation Fails

A derivative may be mathematically correct at the reference case yet give a poor estimate over a wider range. The first-order approximation omits curvature: if the slope changes quickly, the initial derivative will not represent the rate of change across the interval. The approximation can also be inadequate when the outcome is not differentiable at the reference point, as can occur with a hard threshold or a discontinuous rule. A binary recommendation may jump from one category to another even when the underlying score changes smoothly.

Consider a predicted probability $P(x)=100/[1+\exp(-x)]$, in percentage points, where $x$ is a hypothetical model input on a log-odds scale. At $x=0$, $P=50$ and $dP/dx=100(0.5)(1-0.5)=25$ percentage points per unit. The linear approximation at $x=2$ is $50+25(2)=100$ percentage points. But direct evaluation gives $100/[1+\exp(-2)]=88.08$ percentage points. The 11.92-point difference shows why a derivative at the reference case should not be extended across a wide range without checking the model.

Useful checks include evaluating the outcome at nearby points on both sides of the reference case, comparing the linear approximation with direct model evaluations, and inspecting whether the slope changes over the range of interest. A central finite difference,

\left.\frac{\partial f}{\partial\theta_j}\right|_{\boldsymbol{\theta}_0} \approx \frac{f(\boldsymbol{\theta}_0+h\mathbf{e}_j)-f(\boldsymbol{\theta}_0-h\mathbf{e}_j)}{2h},

can be used to check an analytic derivative when $h$ is sufficiently small and both evaluations are defined. A very large $h$ averages behavior over a wider interval rather than checking the local slope; an extremely small $h$ can make numerical round-off important. If the finite-difference estimate changes substantially across reasonable step sizes, investigate the implementation, scale, boundary behavior and curvature before interpreting the derivative.

Local derivatives complement, rather than replace, other sensitivity approaches. One-way sensitivity analysis examines changes over selected input ranges; threshold plots focus on where an outcome crosses a specified boundary; global methods such as Sobol indices summarize sensitivity across input distributions. A local derivative answers a different question: the rate of change at one reference case. It does not rank inputs across their full plausible ranges, account by itself for uncertainty in those inputs, or establish that a change in a model parameter causes a clinical effect.

Common Pitfalls

Ignoring the reference case. A derivative is conditional on where it was evaluated. State the reference values along with the derivative.
Dropping units or input scaling. A numerical magnitude without units can be misleading, especially when parameters use different scales.
Treating a partial derivative as a joint scenario change. A partial derivative changes one input while holding the others fixed. For coordinated changes, use the gradient dotted with the specified change vector or evaluate the scenario directly.
Extending a local slope across a wide range. The wrong reasoning is, “The derivative is constant enough because it was calculated precisely at the reference case.” Precision at a point does not establish linearity over an interval. Check direct predictions at the interval’s endpoints or at intermediate values.
Interpreting sensitivity as uncertainty or causation. A large derivative does not mean that the input is highly uncertain, nor does a model derivative establish a causal clinical effect.

Key takeaway: A local derivative is a rate of change of a specified decision outcome at a specified reference case. Interpret its sign and units, use the first-order approximation only for sufficiently small changes, and compare it with direct model evaluations before drawing conclusions over a wider range.

Check Your Understanding

Try each question, then compare your reasoning with the solutions below.

A model gives $f(\theta)=3\theta^2$. At $\theta=2$, calculate the local derivative and use it to estimate the change in $f$ for an increase of 0.1.
If a local derivative is $+0.25$ outcome points per mg, what does its sign and unit say about a small increase in dose at that reference case?
Why might a partial derivative be insufficient when clearance and dose both change together?
A derivative is computed at a reference case and then used to predict an outcome 50% farther along the input scale. What check would help assess whether that prediction is reliable?

Solutions. (1) Since $df/d\theta=6\theta$, the derivative at $\theta=2$ is 12 outcome units per unit of $\theta$. The first-order estimate for $\Delta\theta=0.1$ is $\Delta f\approx12(0.1)=1.2$. (2) At that reference case, a small 1 mg increase in dose corresponds locally to an approximate increase of 0.25 outcome points, with other inputs held fixed. (3) The partial derivative changes one input while holding the other fixed; when both change, their joint first-order effect depends on both derivatives and the sizes and directions of both changes. (4) Evaluate the model directly at the proposed input and, if useful, at intermediate points, then compare those results with the linear approximation. A large discrepancy indicates that the local slope does not adequately represent the wider range.