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@(@\newcommand{\W}[1]{ \; #1 \; } \newcommand{\R}[1]{ {\rm #1} } \newcommand{\B}[1]{ {\bf #1} } \newcommand{\D}[2]{ \frac{\partial #1}{\partial #2} } \newcommand{\DD}[3]{ \frac{\partial^2 #1}{\partial #2 \partial #3} } \newcommand{\Dpow}[2]{ \frac{\partial^{#1}}{\partial {#2}^{#1}} } \newcommand{\dpow}[2]{ \frac{ {\rm d}^{#1}}{{\rm d}\, {#2}^{#1}} }@)@This is cppad-20221105 documentation. Here is a link to its current documentation .
Gradient of Determinant Using LU Factorization: Example and Test
// Complex examples should supppress conversion warnings
# include <cppad/wno_conversion.hpp>

# include <cppad/cppad.hpp>
# include <cppad/speed/det_by_lu.hpp>

// The AD complex case is used by this example so must
// define a specializatgion of LeqZero,AbsGeq for the AD<Complex> case
namespace CppAD {
    CPPAD_BOOL_BINARY( std::complex<double> ,  AbsGeq   )
    CPPAD_BOOL_UNARY(  std::complex<double> ,  LeqZero )
}


bool HesLuDet(void)
{   bool ok = true;

    using namespace CppAD;
    typedef std::complex<double> Complex;
    double eps99 = 99.0 * std::numeric_limits<double>::epsilon();

    size_t n = 2;

    // object for computing determinants
    det_by_lu< AD<Complex> > Det(n);

    // independent and dependent variable vectors
    CPPAD_TESTVECTOR(AD<Complex>)  X(n * n);
    CPPAD_TESTVECTOR(AD<Complex>)  D(1);

    // value of the independent variable
    size_t i;
    for(i = 0; i < n * n; i++)
        X[i] = Complex( double(i), -double(i) );

    // set the independent variables
    Independent(X);

    D[0]  = Det(X);

    // create the function object
    ADFun<Complex> f(X, D);

    // argument value
    CPPAD_TESTVECTOR(Complex)     x( n * n );
    for(i = 0; i < n * n; i++)
        x[i] = Complex( double(2 * i) , double(i) );

    // first derivative of the determinant
    CPPAD_TESTVECTOR(Complex) H( n * n * n * n );
    H = f.Hessian(x, 0);

    /*
    f(x)     = x[0] * x[3] - x[1] * x[2]
    f'(x)    = ( x[3], -x[2], -x[1], x[0] )
    */
    Complex zero(0., 0.);
    Complex one(1., 0.);
    Complex Htrue[]  = {
        zero, zero, zero,  one,
        zero, zero, -one, zero,
        zero, -one, zero, zero,
         one, zero, zero, zero
    };
    for( i = 0; i < n*n*n*n; i++)
        ok &= NearEqual( Htrue[i], H[i], eps99 , eps99 );

    return ok;
}
Input File: example/general/hes_lu_det.cpp