Question

what is d/dx ((x^x)^x)=

Answer 1

NotationDerivatives of Linear ProductsDerivatives of Quadratic Products

Notation

d/dx (y) is a vector whose (i) element is dy(i)/dxd/dx (y) is a vector whose (i) element is dy/dx(i)d/dx (yT) is a matrix whose (i,j) element is dy(j)/dx(i)d/dx (Y) is a matrix whose (i,j) element is dy(i,j)/dxd/dX (y) is a matrix whose (i,j) element is dy/dx(i,j)

Note that the Hermitian transpose is not used because complex conjugates are not analytic.

In the expressions below matrices and vectors A, B, C do not depend on X.

Derivatives of Linear Products

d/dx (AYB) =A * d/dx (Y) * Bd/dx (Ay) =A * d/dx (y)d/dx (xTA) =Ad/dx (xT) =Id/dx (xTa) = d/dx (aTx) = ad/dX (aTXb) = abTd/dX (aTXa) = d/dX (aTXTa) = aaTd/dX (aTXTb) = baTd/dx (YZ) =Y * d/dx (Z) + d/dx (Y) * Z

Derivatives of Quadratic Products

d/dx (Ax+b)TC(Dx+e) = ATC(Dx+e) + DTCT(Ax+b)d/dx (xTCx) = (C+CT)x[C: symmetric]: d/dx (xTCx) = 2Cxd/dx (xTx) = 2xd/dx (Ax+b)T (Dx+e) = AT (Dx+e) + DT (Ax+b)d/dx (Ax+b)T (Ax+b) = 2AT (Ax+b)[C: symmetric]: d/dx (Ax+b)TC(Ax+b) = 2ATC(Ax+b)d/dX (aTXTXb) = X(abT + baT)d/dX (aTXTXa) = 2XaaTd/dX (aTXTCXb) = CTXabT + CXbaTd/dX (aTXTCXa) = (C + CT)XaaT[C:Symmetric] d/dX (aTXTCXa) = 2CXaaTd/dX ((Xa+b)TC(Xa+b)) = (C+CT)(Xa+b)aT

Derivatives of Cubic Products

d/dx (xTAxxT) = (A+AT)xxT+xTAxI

Derivatives of Inverses

d/dx (Y-1) = -Y-1d/dx (Y)Y-1

Derivative of Trace

Note: matrix dimensions must result in an n*n argument for tr().

d/dX (tr(X)) = Id/dX (tr(Xk)) =k(Xk-1)Td/dX (tr(AXk)) = SUMr=0:k-1(XrAXk-r-1)Td/dX (tr(AX-1B)) = -(X-1BAX-1)Td/dX (tr(AX-1)) =d/dX (tr(X-1A)) = -X-TATX-Td/dX (tr(ATXBT)) = d/dX (tr(BXTA)) = ABd/dX (tr(XAT)) = d/dX (tr(ATX)) =d/dX (tr(XTA)) = d/dX(tr(AXT)) = Ad/dX (tr(AXBXT)) = ATXBT + AXBd/dX (tr(XAXT)) = X(A+AT)d/dX (tr(XTAX)) = XT(A+AT)d/dX (tr(AXTX)) = (A+AT)Xd/dX (tr(AXBX)) = ATXTBT + BTXTAT[C:symmetric] d/dX (tr((XTCX)-1A) = d/dX (tr(A (XTCX)-1) = -(CX(XTCX)-1)(A+AT)(XTCX)-1[B,C:symmetric] d/dX (tr((XTCX)-1(XTBX)) = d/dX (tr( (XTBX)(XTCX)-1) = -2(CX(XTCX)-1)XTBX(XTCX)-1 + 2BX(XTCX)-1

Derivative of Determinant

Note: matrix dimensions must result in an n*n argument for det().

d/dX (det(X)) = d/dX (det(XT)) = det(X)*X-Td/dX (det(AXB)) = det(AXB)*X-Td/dX (ln(det(AXB))) = X-Td/dX (det(Xk)) = k*det(Xk)*X-Td/dX (ln(det(Xk))) = kX-T[Real] d/dX (det(XTCX)) = det(XTCX)*(C+CT)X(XTCX)-1[C: Real,Symmetric] d/dX (det(XTCX)) = 2det(XTCX)* CX(XTCX)-1[C: Real,Symmetricc] d/dX (ln(det(XTCX))) = 2CX(XTCX)-1

Jacobian

If y is a function of x, then dyT/dx is the Jacobian matrix of y with respect to x.

Its determinant, |dyT/dx|, is the Jacobian of y with respect to x and represents the ratio of the hyper-volumes dy and dx. The Jacobian occurs when changing variables in an integration: Integral(f(y)dy)=Integral(f(y(x)) |dyT/dx| dx).

Hessian matrix

If f is a function of x then the symmetric matrix d2f/dx2 = d/dxT(df/dx) is the Hessian matrix of f(x). A value of x for which df/dx = 0 corresponds to a minimum, maximum or saddle point according to whether the Hessian is positive definite, negative definite or indefinite.

d2/dx2 (aTx) = 0d2/dx2 (Ax+b)TC(Dx+e) = ATCD + DTCTAd2/dx2 (xTCx) = C+CTd2/dx2 (xTx) = 2Id2/dx2 (Ax+b)T (Dx+e) = ATD + DTAd2/dx2 (Ax+b)T (Ax+b) = 2ATA[C: symmetric]: d2/dx2 (Ax+b)TC(Ax+b) = 2ATCA