= Solution
Differentiate the <eigenvalue equation> along the assumed smooth normalized eigenbranch:
$$
A'x+Ax'=\lambda'x+\lambda x'.
$$
Since $A$ is real symmetric, $x^TA=\lambda x^T$. Multiplying the differentiated equation by $x^T$ and using $x^Tx=1$ cancels the terms containing $x'$:
$$
x^TA'x+\lambda x^Tx'=\lambda'+\lambda x^Tx'.
$$
Thus the finite-dimensional <Hellmann–Feynman theorem> gives
$$
\boxed{\left.\frac{d\lambda}{d\epsilon}\right|_{\epsilon=0}=x(0)^T\left.\frac{dA}{d\epsilon}\right|_{\epsilon=0}x(0).}
$$
Equivalently, differentiating the normalization gives $2x^Tx'=0$ and eliminates those terms from the <derivative> of $\lambda=x^TAx$. The proof needs the smooth eigenbranch assumed in the question; the smallest <eigenvalue> need not be differentiable at an arbitrary crossing without that assumption.
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