derivative of normalizer in exponential form -- change integral and gradient

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When deriving the relation between normalizer and expectation of the sufficient statistic for distributions in exponential form one uses the fact, that the density integrates to one: $$1 = g(\theta)\int h(x) \exp(\eta(\theta)\cdot T(x))dx$$

Gradient wrt to $\theta$ on both sides, chain rule and rearranging yields $$- \nabla g(\theta) \int h(x) \exp(\eta(\theta)\cdot T(x))dx = g(\theta) \nabla\int h(x) \exp(\eta(\theta)\cdot T(x))dx$$

Why is it possible to change the order of gradient and integral at the RHS? I.e. $$g(\theta)\nabla\int h(x) \exp(\eta(\theta)\cdot T(x))dx = \int g(\theta)\nabla (h(x) \exp(\eta(\theta)\cdot T(x)))dx = ET(x)$$

After this rearranging and simplifying yields $$- \frac{\nabla g(\theta)}{g(\theta)} = ET(x)$$

But I struggle to justify why I can take the gradient under the integral here...