Partial derivative multiplied with delta (difference)

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once again I am stuck with a derivation in thermodynamics.

We start from $dS=\frac{dU}{T}+\frac{p \cdot dV}{T}$ and a result in the lecture notes is:

$\Delta(\frac{1}{T}) = \frac{\partial^2S}{\partial^2U}\cdot\Delta U + \frac{\partial^2S}{\partial U\partial V}\cdot\Delta V$.

The intermediate steps seem to be $\frac{\partial S}{\partial U}=\frac{1}{T}$, so $\frac{\partial^2S}{\partial^2U}=\frac{\partial(1/T)}{\partial U}$ and $\frac{\partial^2S}{\partial U\partial V}=\frac{\partial(1/T)}{\partial V}$. It somehow seems reasonable to me to multiply with a difference of V or U to get (1/T), but it doesn't really make sense to me why this would work. Multiplying with a difference can't be the same as integrating, can it? Or is the integration just implicit?