On Thursday, September 8, 2016 at 4:18:42 PM UTC-4, Mathieu Taschereau-Dumouchel wrote: > > Not quite because the x_i on the right-hand side also depend on a_j
If you just take the partial derivative of both sides with respect to aₖ (for k = 1..N), you will see that (given the solution vector x) you just get a *linear* system of equations for the gradient: ∂xⱼ/∂aₖ - b aⱼ ∑ᵢ (xᵢ)ᵇ⁻¹ ∂xᵢ/∂aₖ = δⱼₖ ∑ᵢ (xᵢ)ᵇ Notice that the right-hand side is proportional to the identity matrix (δ denotes the Kronecker delta). So, just form the matrix corresponding to the linear operator multiplying ∇ₐx on the left-hand side, and your Jacobian matrix ∇ₐx will then be the inverse of this matrix multiplied by ∑ᵢ (xᵢ)ᵇ.
