fluid dynamics - Derivation of Kelvin's circulation theorem

In the derivation of Kelvin's circulation theorem, I take the material derivative of circulation, or

$$\begin{align} \dfrac{D\Gamma}{Dt} = \dfrac{D}{Dt}\oint_C \vec{u} \cdot d\vec{\ell}. \end{align}$$

Moving the material derivative inside the integral gives

$$\begin{align} \dfrac{D\Gamma}{Dt} = \oint_C \dfrac{D\vec{u}}{Dt} \cdot d\vec{\ell} + \oint_C \vec{u} \cdot \dfrac{D(d\vec{\ell})}{Dt}. \end{align}$$

Given that $d\vec{\ell} = ds \hspace{1mm} \vec{t}$, the second integral of the equation above becomes

$$\begin{align} \oint_C \vec{u} \cdot d\vec{u}. \end{align}$$

How does one arrive at this equation? My attempt would be to use the chain rule so that

$$\begin{align} \dfrac{D(ds \vec{t})}{Dt} = ds\dfrac{D\vec{t}}{Dt} + \vec{t}\dfrac{D(ds)}{Dt}. \end{align}$$

From here, I expand each material derivative as

$$\begin{align} \dfrac{D\vec{t}}{Dt} = \dfrac{\partial \vec{t}}{\partial t} + u_t\dfrac{\partial \vec{t}}{\partial s}+ u_n\dfrac{\partial \vec{t}}{\partial n}\\ \dfrac{D(ds)}{Dt} = \dfrac{\partial ds}{\partial t} + u_t\dfrac{\partial ds}{\partial s}+ u_n\dfrac{\partial ds}{\partial n}. \end{align}$$

I am unsure how these two relations will simplify so that I get

$$\dfrac{D(d\vec{\ell})}{Dt} = d\vec{u}.$$