 Solution

Differentiate the eigenvalue equation along the assumed smooth normalized eigenbranch:Since $A$ is real symmetric, $x^{T}A=\lambda x^T$. Multiplying the differentiated equation by $x^T$ and using $x^{T}x=1$ cancels the terms containing $x^{\prime }$:Thus the finite-dimensional Hellmann–Feynman theorem givesEquivalently, differentiating the normalization gives $2x^T{x}^{\prime }=0$ and eliminates those terms from the derivative of $\lambda =x^{T}Ax$. The proof needs the smooth eigenbranch assumed in the question; the smallest eigenvalue need not be differentiable at an arbitrary crossing without that assumption.