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Derivative of the Abs Function in Mathematica looks very different than the actual answer. Why? How to correct this?

For example I entered D[Abs[x - 1], x] and got the output Abs'[-1 + x]. I was hoping to get the output in this form: (x−1)/|x−1|.

Is there a syntax error in my input?

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  • $\begingroup$ FullSimplify[D[Abs[x - 1], x], x \[Element] Reals && x != 0] Resolve[ForAll[x, x \[Element] Reals && x != 0, Sign[-1 + x] == Piecewise[{{1, x - 1 > 0}, {-1, x - 1 < 0}}]], Reals] $\endgroup$
    – ciao
    Commented May 1, 2014 at 7:29
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    $\begingroup$ How do I get the output in this form: (x−1)/|x−1| ? $\endgroup$
    – user14056
    Commented May 1, 2014 at 7:34
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    $\begingroup$ Throw a rule and traditionalform onto end of simplify, /. Sign[z_ - 1] :> (z - 1)/Abs[z - 1] // TraditionalForm or investigate ComplexityFunction and TransformationFunctions for simplify. Bottom line, the form presented is what Mathematica deems the "simplest". If you need/want to transform it, you need to tell it so. $\endgroup$
    – ciao
    Commented May 1, 2014 at 7:41
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    $\begingroup$ You need ComplexExpand probably because you can't take the derivative of Abs in the complexes $\endgroup$
    – Rojo
    Commented May 1, 2014 at 15:17
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    $\begingroup$ Well, Abs[x-1] is in fact not differentiable in the classical sense at x=1. So x-1/|x-1| would only be the right answer for x != 1. $\endgroup$
    – Wizard
    Commented May 1, 2014 at 17:21

2 Answers 2

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In version 11.1 the function RealAbs was introduced that give your desired result

D[RealAbs[x - 1], x]
RealAbs'[x - 1]

(*(-1 + x)/RealAbs[-1 + x]*)
(*(-1 + x)/RealAbs[-1 + x]*)

Alternatively, as suggested by @murray, you can use ComplexExpand

D[Abs[x - 1], x] // ComplexExpand // Together
(*(-1 + x)/Sqrt[(-1 + x)^2]*)

Or explicitly state that $x$ is real, as @Rojo said

D[Abs[x - 1], x] // FullSimplify[#, x \[Element] Reals] &

(*Sign[-1 + x]*)

However, FullSimplify changes the domain slightly because Sign[-1 + x] is defined at $x=1$

Sign[x - 1] /. x -> 1
(*0*)

RealAbs'[x - 1] /. x -> 1
(*Indeterminate*)
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  • $\begingroup$ RealAbs'[x - 1]==Sign[x - 1] are equal.Check: Limit[{RealAbs'[x - 1], Sign[x - 1]}, x -> 1, Direction -> #] & /@ {-1, 1} $\endgroup$ Commented Jun 3, 2018 at 17:17
  • $\begingroup$ @MariuszIwaniuk their limits are equal at x=1, but try this {FunctionDomain[RealAbs'[x - 1], x], FunctionDomain[Sign[x - 1], x]} $\endgroup$
    – user45937
    Commented Jun 4, 2018 at 1:54
  • $\begingroup$ {FunctionDomain[ Simplify[RealAbs'[x - 1], Assumptions -> {x \[Element] Reals}], x], FunctionDomain[Sign[x - 1], x]} ? $\endgroup$ Commented Jun 4, 2018 at 9:14
  • $\begingroup$ @user45937 Hi. D[Abs[x - 1], x] // ComplexExpand // Together failed in v13. $\endgroup$
    – lotus2019
    Commented Jun 21, 2023 at 23:12
  • $\begingroup$ @lotus2019 in v13.2.1 it returns (-1 + x)/Sqrt[(-1 + x)^2] $\endgroup$
    – user45937
    Commented Jun 22, 2023 at 16:57
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It has been mentioned before (for example, see this answer) that Abs in Mathematica is defined for complex numbers. Since Abs is not holomorphic over the complex numbers, its derivative is not well-defined. One way to see this is:

FullSimplify[Abs[z] == Sqrt[z Conjugate[z]]]

True

Here are a couple more ways to achieve what you want (besides those mentioned by @roman).

  1. Use Sqrt[z^2] instead of Abs[z]:

D[Sqrt[z^2], z]

z/Sqrt[z^2]

  1. Use complex (Wirtinger) derivatives. Summarizing:

$$\frac{\partial}{\partial z} = \frac{1}{2} \left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right), \quad\frac{\partial}{\partial \bar{z}} = \frac{1}{2} \left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right)$$

This means that the "real" derivative, i.e., the derivative with respect to $x$ only, is given by:

$$\frac{\partial}{\partial x} = \frac{\partial}{\partial z} + \frac{\partial}{\partial \bar{z}}$$

Hence the "real" derivative of Abs[z] is:

$$\frac{\partial}{\partial x}\left| z \right| = \frac{\partial}{\partial x} \sqrt{z \bar{z}} = \frac{z + \bar{z}}{2 \left| z \right|} = \frac{\Re{z}}{\left| z \right|}$$

which agrees with the previous answer when $z$ is real.

However, this exposition is not useful without an actual implementation of Wirtinger derivatives! On the other hand, in this answer I give such an implementation, and the function is called ComplexD (I provide its definition at the end of this answer). So:

ComplexD[Abs[z], z] + ComplexD[Abs[z], Conjugate[z]]
% //FullSimplify

z/(2 Abs[z]) + Conjugate[z]/(2 Abs[z])

Re[z]/Abs[z]

in agreement with the above results. It would be possible to define a RealD function:

RealD[expr_, z_] := ComplexD[expr, z] + ComplexD[expr, Conjugate[z]]

if one were so inclined. Here is the definition of ComplexD (slightly edited from the original):

ComplexD[expr_, z__] := With[
    {
    nc = NonConstants -> Union @ Cases[{z},
        s_Symbol | Conjugate[s_Symbol] | {s_Symbol | Conjugate[s_Symbol], _} :> s
    ],
    old = OptionValue[
        SystemOptions[],
        "DifferentiationOptions" -> "ExcludedFunctions"]
    },

    Internal`WithLocalSettings[
        With[{new = Join[old, {Abs, Conjugate}]},
            SetSystemOptions["DifferentiationOptions"->"ExcludedFunctions" -> new]
        ];
        Unprotect[Conjugate, Abs];
        Conjugate /: D[w_, Conjugate[w_], nc] := 0;
        Conjugate /: D[Conjugate[f_], w_, nc] := Conjugate[D[f, Conjugate[w], nc]];
        Abs /: D[Abs[f_], w_, nc] := D[Conjugate[f]f, w, nc]/(2 Abs[f]),

        D[expr, z, nc],

        SetSystemOptions["DifferentiationOptions" -> "ExcludedFunctions" -> old];
        Conjugate /: D[w_, Conjugate[w_], nc] =.;
        Conjugate /: D[Conjugate[f_], w_, nc] =.;
        Abs /: D[Abs[f_], w_, nc] =.;
        Protect[Conjugate, Abs];
    ]
]
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