Under a Poincare transformation $$x^{\prime\mu}=x^\mu+\delta x^\mu=\Lambda^{\mu}{}_{\nu}x^\nu+\epsilon^\mu.\tag{1}$$ where first term corresponds to an arbitrary Lorentz transformation and the second term represents a translation in spacetime. I'm trying to calculate the Jacobian of this transformation $J=\det\Big(\frac{\partial x^{\mu\prime}}{\partial x^\lambda}\Big)$. Differentiating (1) we get, $$\frac{\partial x^{\prime\mu}}{\partial x^\lambda}=\Lambda^{\mu}{}_{\nu}\delta^{\nu}{}_\lambda+\frac{\partial \epsilon^\mu}{\partial x^\lambda}=\Lambda^{\mu}{}_{\lambda}+\partial_\lambda\epsilon^\mu\tag{2}$$ where $\partial_\lambda$ is the shorthand for $\frac{\partial}{\partial x^\lambda}$.
Next, I'm supposed to take the determinant on both sides. But the determinant of the sum of two matrices A and B is not equal to the sum of the determinants i.e., $$\det(A+B)\neq \det(A)+\det(B).$$ So how do I evaluate $J=\det\Big(\frac{\partial x^{\mu\prime}}{\partial x^\lambda}\Big)$?
Ryder's book on Quantum field theory use the first equality of (1), and write $$\det\Big(\frac{\partial x^{\mu\prime}}{\partial x^\lambda}\Big)=1+\partial_\mu(\delta x^\mu).\tag{3}$$ I have no clue how they got (3). This is not a typo because the same result is also used in Walter Greiner's Field Quantization Eq. 2.47.