Suppose a function $f(x): \mathbb R^d \mapsto \mathbb R^D$, and its stochastic approximator, $g(x; W): \mathbb R^d \mapsto \mathbb R^D$. Here $W$ is some random variable. Then $g(x; W)$ is unbiased in the sense that $$\mathbb E_W [g(x;W)] = f(x),$$ for any $x$.
I think the following two are not equal, but how to prove it? $$\mathbb E_W\left[\frac{dg(x;W)}{dx}\right] \text{ vs. } \frac{df(x)}{dx}.$$
Indeed, the equality will not hold in general. For counterexamples, see this or this.
For sufficient conditions for the equality when $d=1$, see e.g. Folland, Theorem 2.27 or the more general Lemma 2.3. This immediately extends to any $d$ if the derivatives are understood in the Gateaux sense. As seen from the discussion of Theorem 2.27 in Folland, the extension to $d>1$ is somewhat more problematic if the derivatives are understood in the Fréchet sense. This answer may also be of interest to you.
The two expressions are equal, because the operations of taking an expectation with respect to a random variable $W$ and taking a derivative with respect to a parameter $x$ commute: $$\mathbb E_W\left[\frac{dg(x;W)}{dx}\right] = \lim_{\delta\rightarrow 0}\delta^{-1}\bigl(\mathbb E_W[g(x+\delta;W)]-E_W[g(x;W)]\bigr)$$ $$=\lim_{\delta\rightarrow 0}\delta^{-1}\bigl(f(x+\delta)-f(x)\bigr)=\frac{d}{dx}f(x).$$
