[Creatinine] can change in an unexpected direction due to the volume change rate that interacts with kinetic GFR: Potentially positive paradox

Abstract [Creatinine] was proved to change in the opposite direction of the kinetic GFR (GFRK), but does the [creatinine] also change in the opposite direction of the volume rate? If volume is administered and the [creatinine] actually goes up, then the two changes move in the same direction and their ratio is positive, paradoxically. The equation that describes [creatinine] as a function of time was differentiated with respect to the volume rate. This partial first derivative has a global maximum that can be positive under definable conditions. Knowing what makes the maximum positive informs when the derivative will be positive over some continuous domain of volume rate inputs. The first derivative versus volume rate curve has a maximum and a minimum point depending on the GFRK. If GFRK is below a calculable value, then the curve's minimum vanishes, letting it descend to ‐∞ and not allowing the derivative to ever be positive. If GFRK lies between a lower and a higher calculable value, then the curve's maximum vanishes, letting the derivative diverge to +∞, though the clinical scenario is unrealistic. If GFRK is above the higher calculable value, then the curve's absolute maximum can become positive by decreasing the creatinine generation rate or increasing the initial [creatinine]. The derivative is potentially positive under these clinically realizable circumstances. The combination of parameters above can align in septic patients (low creatinine generation rate) with kidney failure (high initial [creatinine]) who are put on continuous dialysis (high GFRK). If a first derivative is positive, removing more volume can improve the [creatinine] and, dismayingly, giving more volume can worsen the [creatinine]. This paradox is explained by a covert interplay between the ambient [creatinine] and GFRK that excretes creatinine faster than its volume of distribution declines.


| INTRODUCTION
We previously showed that changes in the glomerular filtration rate (GFR) must always drive the serum creatinine in the opposite direction (Chen & Chiaramonte, 2021). If the GFR were to decrease, then the creatinine would have to increase, and vice versa. Though these statements seem obvious, they were only recently proved by differentiating the creatinine with respect to kinetic GFR (Chen, 2013) and finding that this partial derivative's sign is always negative for any possible set of real-world values. It does not matter how extreme the variables are or how they are combined. The perpetual negative sign assures doctors that a rise in creatinine (positive Δcreatinine) had to come about from a drop in kinetic GFR (negative ΔGFR K ) and vice versa, with all else being constant-particularly time.
Is creatinine also related to the volume change rate in an ever-opposing way? If more volume is being gained (positive Δvolume rate), the [creatinine] is further diluted and Δ [Cr] is negative. If more volume is being lost (negative Δvolume rate), the [creatinine] is further concentrated and Δ [Cr] is positive. With the signs being opposite in our thought experiments, it would appear that the partial derivative of [Cr] with respect to volume change rate is always negative, same as for kinetic GFR (Chen & Chiaramonte, 2021). But could unsuspected factors alter the sign? What if a negative Δvolume rate concentrates the [creatinine] even further, but that enables the kinetic GFR to excrete more creatinine? Would the creatinine quantity decline faster than the volume of distribution, making the creatinine concentration fraction lower in value-a negative Δ [Cr]? If so, the partial derivative would be positive, with volume rate and [Cr] decreasing over the same time frame. The potential for a positive sign brings up a clinical paradox. Sometimes, being more aggressive with the volume removal may improve the serum creatinine (Hegde, 2020). Or, giving even more volume may increase the creatinine. These paradoxes can occur in septic patients on continuous dialysis, for purely mathematical reasons to be shown that need not involve the messiness of real life, which is more complex than the derivative model that assumes that only volume and [creatinine] can change.

| Creatinine kinetics
The differential equation that underpins the kinetic GFR states that the rate of change in the creatinine mass is equal to the creatinine input rate minus the creatinine output rate (Chen, 2018a(Chen, , 2018bChen & Chiaramonte, 2019). Further, the creatinine mass at any given time is the current [creatinine] times the volume of distribution, typically taken to be total body water (TBW) (Bjornsson, 1979;Chow, 1985;Edwards, 1959;Jones & Burnett, 1974;Pickering et al., 2013). To account for the concentrating and diluting effects on the [creatinine], its volume of distribution can be modeled to change at a constant rate: V t = V 0 + ΔV Δt t, where V t is the volume as a function of time, V 0 is the initial volume, ΔV Δt is the (average) volume change rate, and t is time. The creatinine input rate is primarily determined by the muscle mass, which tends to be fairly stable so that the creatinine generation rate is usually thought of as a constant: Gen. The creatinine output rate is mostly determined by the kidney such that the excretion rate is equal to the kinetic GFR times the ambient [creatinine]: where [Cr] t is the [creatinine] at a particular time. Thus, the differential equation is: This first-order linear differential equation's solution, as previously published, is (Chen, 2018a): In other words, the serum [creatinine] at a given time is equal to the initial [creatinine] [Cr] 0 plus a time-evolved portion of the spread between the initial [creatinine] and the [creatinine] reached at a new steady state if the kinetic GFR and volume change rate remained at those levels.

| Derivative of [creatinine] with respect to volume change rate
From Equation (2), we can deduce how the serum creatinine would change if one other variable were tweaked, and the partial derivative is suited to this task. Previously, the one other variable was kinetic GFR (Chen & Chiaramonte, 2021), but now the one other variable will be the volume change rate. The derivative of [Cr] t with respect to ΔV Δt quantifies their relationship at every instant, allowing a comprehensive assessment of the sign. If the sign can be positive, then [Cr] t may change in the same direction as ΔV Δt . ( In Equation (2), the derivative of ⎡ with respect to ΔV Δt is: To calculate the derivative of the exponential, let and use logarithms.
Differentiate with the product rule: Next, in Equation (2), find the derivative of

| Calculator and concept map
To follow the calculations in the Results, please download a spreadsheet we created to calculate the main equations in the manuscript. You can use the spreadsheet to explore your own scenarios and questions. For a map of the concepts being presented, the final algorithm in Section 3.8 may help with understanding when the first derivative in Equation (5) can be positive. First, Gen and [Cr] 0 will be varied (Section 3.5), as their ratio is a principal determinant of positivity. Later, GFR K will also be varied (Section 3.6), as values above a calculable reference point can allow the first derivative to be positive in situations that are clinically encountered.

| First derivative behavior and sign
To gauge the behavior and sign of the partial first derivative, we graphed

ΔV Δt
(y-axis) vs. ΔV Δt (x-axis) for an acute Thus, kidney injury (AKI): steady state GFR of 100 ml/min corresponding to an initial [Cr] of 1.0 mg/dl that increases over the next 24 hours when the GFR K suddenly drops to 20 ml/min in a patient with a TBW (volume of distribution) of 42 L. According to the thought experiments, the sign of

ΔV Δt
should be negative (below y = 0 ) throughout the gamut of ΔV Δt values ( Figure 1). As ΔV Δt approaches an extreme that would deplete all of the TBW by the 24-h mark (in this case), the [Cr] t ΔV Δt goes to − ∞ (Figure 1). A volume of zero is nonsensical, so if V 0 + ΔV Δt t needs to be > 0 , then become positive under certain conditions? To find out, we varied the parameters and found conditions that work: creatinine generation rate (Gen) of 60 mg/dl × ml/min and [Cr] 0 of 8.0 mg/dl that decreases over the next 24 h when the GFR K suddenly increases to 100 ml/min (e.g., by renal replacement therapy) in a patient with a TBW of 42 L. The is more likely to be positive if Gen is small and [Cr] 0 is high; it helps if GFR K is larger and ΔV Δt is negative. This family of curves has an absolute maximum. If we can find the curve whose maximum lies tangent to y = 0, that represents the border between a first derivative being perpetually negative versus potentially positive. In Figure 2, the Gen = 70 (blue) curve comes closest to touching y = 0. Its maximum

ΔV Δt
is − 0.0008, but we can place the peak right at 0. F I G U R E 1 Example first derivative of [creatinine] with respect to volume change rate. Equation (5) is graphed with ΔV Δt as an independent variable (x-axis) and flattens out as it approaches zero asymptotically. The first derivative is always negative in this example of acute kidney injury

| First derivative's peak
To calculate the peak of the = 0 and solved for ΔV Δt by Newton's method or the secant method. At one ΔV Δt root, our example first derivative (Section 3.1, second paragraph) attains its maximum and is positive. On either side of the ΔV Δt root, the  becomes a complex number: in Equation (5) , one that happens to be absolute, and a sin-

ΔV Δt
, one that is relative.

| Making the first derivative's peak tangent to the x-axis
Setting the second derivative equal to zero optimizes the first derivative, but the first derivative's absolute maximum is not necessarily zero. To find a curve whose maximum is tangent to y = 0, we devised a way to make both the first derivative and the second derivative equal to zero at the same time. In doing so, we find the transition to In the first and second derivatives [Equations (5) and (6)], only two variables can be explicitly solved for, namely Gen and [Cr] 0 . Set the first derivative equal to zero and solve for [Cr] 0 : [Cr] 0 = Gen  ratio is a fixed attribute for a set of GFR K , V 0 , and t inputs that allows the first and second derivatives to equal zero simultaneously.

| Gen and [Cr] 0 effects: lifting the peak into positive territory
Now that the peak can be positioned at the xaxis, how can the [Cr] t ΔV Δt be lifted above the xaxis? The GFR K , V 0 , and t are initial data, and ΔV Δt is the independent variable on the graph. That leaves only Gen and [Cr] 0 to be manipulated.
Using the fixed

Gen
[Cr] 0 ratio as a benchmark, we find that lower ratios shift the curve partially into positive territory, in keeping with the observation that smaller Gens and/or bigger [Cr] (7), for example, and then ask if the patient's actual [Cr] 0 is larger, which lets rises from a negative value to peak at an absolute maximum which can be positive, then falls to a relative minimum (mentioned in Section 3.2) that is negative, and then asymptotically increases toward y = 0. The curve is shifted vertically, more or less, by varying the Gen or [Cr] 0 . Well, the curve is shifted horizontally, mostly, by varying the GFR K . A higher GFR K pulls the curve rightward, and a lower GFR K pushes it leftward. Also, imagine that the left end of the curve is tethered to an invisible wall at ΔV Δt = − V 0 t but has the ability to slide up or down that wall. Then, a right shift would stretch the curve, flattening it out, and a left shift would compress the curve, bunching it up against the wall in an orderly way by making it bend and stack in layers (with no thickness). Can the GFR K be lowered sufficiently to left-shift the absolute maximum until it is located at the leftmost ΔV Δt , that is, − V 0 t ? Going further, can the left shift continue until the relative minimum is then pressed up against the leftmost ΔV Δt wall? If so, these max/min at the leftmost ΔV Δt would correspond to a second derivative equaling zero at two GFR K roots, one for the max and one for the min.
As GFR K is reduced, the curve acts like a rope being pushed leftward against a wall, based on tracking the maximum and minimum [Cr] t ΔV Δt points and the sliding along the wall. In response to the push, the endpoint at the leftmost ΔV Δt moves down, the maximum moves up, and the minimum moves down, like how a rope could fold to be more compact (Figure 4a). In addition to the vertical motions, the max/min points move horizontally to the left. Once the GFR K is lowered to ~58.34 (in this example), the bend at the maximum is very sharp and the maximum is left-shifted all the way to ΔV Δt ≅ − V 0 t ( Figure 4b). As the GFR K is lowered some more, the minimum continues to move down and left but the absolute maximum is transitioned into the left endpoint of the curve sliding up the wall, on its way to + ∞ (Figure 4c). In this way, certain GFR K s can enforce a positive

| GFR K effect: keeping [Cr] t V t negative
As GFR K is further reduced, with no more sliding down the wall for now, the relative minimum becomes an absolute minimum ( Figure 4c). As the GFR K reduction keeps pushing the curve/rope to the left against a wall, the bend gets sharper and the minimum moves even more to the left and down. When the GFR K gets down to ~36.67 (in this example), the minimum is left-shifted all the way to the leftmost ΔV Δt (Figure 4d), like the maximum was earlier. As GFR K is lowered past ~36.67, the absolute minimum is transitioned into the left endpoint of the curve sliding down the wall, on its way to − ∞ (Figure 4e). After this transition, the [Cr] t ΔV Δt will always be negative.
For details on how the kinetic GFR can alter the shape of the first derivative curve and help determine whether  is to compile the lessons above into an algorithm. If the GFR K roots are in a "permissive" order of Gen

ΔV Δt
to be positive, then: F I G U R E 4 Decreasing GFR K pushes the first derivative curve leftward along the ΔV Δt xaxis. Variables in common are Gen = 60 mg/dL⋅ mL/min, [Cr] 0 = 8 mg/dL, V 0 = 42 L, and t = 24 h. (a) As the GFR K decreases from 90 (red) to 80 (blue) to 70 (green) mL/min, the curve looks like it is being pushed to the left and is bending in the process. The maximum moves steadily up, the minimum moves down, and both of them move to the left. Also, the left endpoint slides down a virtual wall at the leftmost ΔV Δt . (b) When GFR K decreases to 58.34 ≅ 2 V 0 t , the maximum has been pushed to the leftmost ΔV Δt , and only a short tail to the left of the maximum is decreasing before it gets truncated at the wall (inset). (c) As GFR K decreases below 58.34, the maximum vanishes (blue) and transitions into a left tail that blows up to + ∞ (red). (d) With no more maximum, the minimum is the sole critical point, and it continues to move down and left as the GFR K decreases further from 57 (red) to 50 (orange) mL/min. When the GFR K drops to 36.67 ≅ Gen [Cr] 0 + V 0 t , the minimum has been pushed to the leftmost ΔV Δt , and the left tail still diverges to + ∞ (purple). (e) When GFR K decreases below 36.67, the minimum vanishes (purple) and transitions into a left tail that plunges to − ∞ (red). From here, the first derivative is always negative a. For 0 ≤ GFR K < Gen (iii) Alternatively, plug the ΔV Δt from step c., i. and a known [Cr] 0 into Equation (9) to calculate a benchmark Gen (see Section 3.5).

If the patient's Gen is less than the benchmark
Gen, then the absolute maximum lies above the xaxis and  (Doi et al., 2009;Prowle et al., 2014). Sepsis may have also caused kidney failure, so the [creatinine] went fairly high. Doctors then initiated continuous renal replacement therapy (CRRT) that provided a GFR K greater than 2 V 0 t . (GFR K here is not used in the literal sense of clearance done by the glomerulus. Rather, it is used in the broader sense of clearance done by any means, including extracorporeal).

| Paradox by the numbers
The abstract math may be easier to grasp if we put some concrete numbers on it. Suppose that a septic patient now has a Gen = 40 mg/dl × ml/min. He develops acute tubular necrosis and the creatinine rises to 8 mg/dl. CRRT is started, and the total GFR K = 80 ml/min. The combination of conditions seems ripe for a positive  (7) to calculate a benchmark [Cr] 0 of 4.08…) The patient's Gen of 40 is less than the benchmark Gen, so the absolute maximum lies above the xaxis. (Alternatively, the patient's [Cr] 0 of 8 is greater than the benchmark [Cr] 0 , and again the absolute maximum lies above the xaxis). If the maximum is positive, then

| Effect size
Say that the CRRT ultrafiltration (UF)-volume removal-rate is turned up from 100 to 300 ml/h, that is, the ΔV Δt goes from − 0.1 to − 0.3 L/h, making the ΔV Δt negative. At those two ΔV Δt s, the first derivative is positive (≈ 0.012 to 0.029 mg/dl per L/h). That forces the [Cr] t to be negative. By Equation (2) . Importantly, the positive sign assures the nephrologist that turning up the UF rate will actually improve the next day's [creatinine]. One might posit that the [Cr] t improvement is due to the higher UF rate increasing convective clearance (Tandukar & Palevsky, 2019), but the math disproves that by holding the GFR K constant. Besides, turning up the UF rate will worsen the next day's [creatinine] if the

| Come-from-behind win: getting to a lower [creatinine]
By itself, volume loss should concentrate and thereby increase the [creatinine]. Somehow, this concentration effect is overridden by a creatinine-lowering effect. In 24 h, ΔV Δt = − 0.3 L/h got to a lower [creatinine] than ΔV Δt = − 0.1 L/h. Having ΔV Δt = − 0.3 L/h would seem like a handicap, because removing more volume concentrates the [creatinine] and resists the GFR K that is trying to lower the [creatinine]. Thus, the ΔV Δt = − 0.3 ( Figure 5, blue curve) has a higher [creatinine] than the ΔV Δt = − 0.1 (Figure 5, red curve) at almost all time points. After about 5.7 h, however, the blue curve starts to catch up to the red curve ( Figure  5), which is peculiar as the two ΔV Δt s have not changed. Apparently, concentrating the [creatinine] can be advantageous when the higher [Cr] t interacts with the steady GFR K to excrete more creatinine mass. That lowers the total creatinine (numerator) faster than its volume (denominator), such that the creatinine concentration starts to decline more quickly. The blue curve catches up to the red curve at ~22 h ( Figure 5). Then, the blue curve barely edges out the red curve at the 24-h mark ( Figure 5, see inset), meaning that the higher UF rate ( − 0.3) came from behind to get to a lower [creatinine]. Despite the concentration disadvantage for most of the race, the higher UF rate's latent factor that slowly predominated was a synergy between the [Cr] t and the GFR K to boost creatinine excretion.

| Volume gain can increase the [creatinine]
In the same clinical example, the Because the UF has a higher [Cr] t that is subjected to a relatively high GFR K for most of the race, more creatinine is excreted that eventually lowers the [creatinine] further versus a gain of volume, even with the latter's dilution effect advantage. So, the creatinine-lowering effect that overcomes the volume effect is facilitated by a higher GFR K , which explains why the GFR K should be > 2 Most patients will not be at risk for a positive paradox. The combination of low Gen, high [Cr] 0 , and high-ish GFR K is rare. The GFR K is < Gen t in many instances of AKI, so those patients are protected from a paradox and likely will behave as expected in response to fluids or diuresis. Milder cases of AKI can have a GFR K that lies between Gen that was proved to always be negative (Chen & Chiaramonte, 2021).

ACKNOWLEDGMENTS
The graphs were created at desmos.com/calculator. The authors thank Amy Wang for technical assistance with the graphs.

CONFLICT OF INTEREST
The authors have no conflicts of interest.

AUTHOR CONTRIBUTION
Sheldon Chen: conception and design of study, mathematical derivations and equation graphs, interpretation of data, writing and revision of manuscript, and final approval of the manuscript. Robert Chiaramonte: confirmation of mathematical derivations and equation graphs, interpretation of data, revision, and final approval of the manuscript. The two roots encoded by our GFR K formulas are the farthest left that a root pair can go, because they represent the GFR K s small enough to shift a first derivative's maximum and then minimum all the way to the leftmost ΔV Δt ( Figure  A1, red). For a first derivative's max/min to be located at any ΔV Δt xcoordinate to the right of − V 0 t , the second derivative's GFR K root pair is going to be larger. To demonstrate, we graphed the second derivative like before but changed the ΔV Δt from − 1.74999 to − 1.7. This plot has less vehement swings, without all of the leftmost ΔV Δt causing division by zero. As predicted, the GFR K root pair for ΔV Δt = − 1.7 is larger and to the right of the 58.34 and 36.667 pair ( Figure  A1, blue dotted curve). If we could shift the GFR K root pair to the left of (36.67, 58.34), then the smaller root would vanish behind a wall at 36.67, turning the pair into a single. That is why, in a sense, that GFR K s between 36.67 and 58.34 lack their partner. That suddenly-single root relates to the first derivative having only a minimum when 36.67 < GFR K < 58.34 (Figure 4c). Further, if we could shift the GFR K root pair to the left of 36.67, then both roots would vanish, forecasting that

F I G U R E A 1
Second derivative goes to infinity near the leftmost ΔV Δt but still has two GFR K roots. Equation (6) is graphed with GFR K as an independent variable (x-axis) and and − 1.7 L/h. For ΔV Δt at its leftmost, the second derivative is volatile. It swings to and from ±∞ (red curve). Yet, it crosses the xaxis twice and therefore has two roots that can be estimated as GFR K ≈ Gen [Cr] 0 + V 0 t = 36. 6 and GFR K ≈ 2 V 0 t = 58. 3 . As ΔV Δt is increased, the second derivative curve shifts to the right (blue dot curve). It is not as volatile, and its roots are greater than when ΔV Δt is leftmost. In a way, the GFR K roots are a continuum, beginning at Gen The two GFR K landmarks divide the GFR K number line into three domains. The top domain is GFR K > 2 V 0 t , like the GFR K of 100 that was chosen for all of the curves in Figures 2 and 3. GFR K s this large allow both the absolute maximum and the relative minimum to be to the right of ΔV Δt = − V 0 t ( Figure A2, blue). If GFR K descends into the middle domain of 2 t , then the graph has a generic shape like the blue curve. It has an absolute maximum that can be positive, as demonstrated by the blue curve. A small part of the curve is positive, and the rest of the curve is negative in terms of the [Cr] t ΔV Δt sign absolute maximum into the left tail of a curve that blows up to + ∞. We graphed an example of this curve shape using GFR K = 52 ml/min, which is between 58.34 and 36.67 ( Figure A2, red).
In special cases, the left tail may remain negative. If the two GFR K roots are so close that the middle domain is squeezed, then an absolute minimum is forced to be near the leftmost ΔV Δt , a wall that truncates the left tail before F I G U R E A 3 Types of derivative curves when the GFR K roots are reversed. (a) Unlike in Figure A2, the Gen Δt is barely increased to − 1.748 (blue), the roots get closer together instead of shifting as a spaced pair to the right, like in Figure A1. Increase ΔV Δt to − 1.747 (green) and the roots get closer still, while staying between 58.343 and 66.643. When ΔV Δt is increased to − 1.746 (purple), the second derivative no longer has roots. (b) The absence of second derivative roots under most conditions means that the first derivative usually lacks a maximum and a minimum when the GFR K roots are reversed. For GFR K > 66.643 , the typical curve approaches y = 0, and the first derivative is always negative (blue). For 66.643 > GFR K > 58.343, the curve has an absolute minimum at ΔV Δt = − 1.748 ≅ − V 0 t , and the first derivative is always negative (red). For 58.343 > GFR K , the curve has no maximum or minimum, again, and the first derivative is always negative (green)