Linearization of a tensor?

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Suppose that $(M,g)$ is a Riemannain manifold and $\nabla$ denotes its Levi-Civita connection. Define a‎ $(0,2)‎$ tensor field $\pi$ by‎ $$‎\pi (X,Y)=(‎\nabla_X \omega‎)(Y)-\omega (X)\omega (Y)+‎\dfrac{1}{2}‎\omega (‎U‎)‎g(X,Y)‎‎$$ where $\omega$ is an 1-form on $(M,g)$ and $U$ is its equivalent vector field, in fact for all $X\in \mathcal{X}(M)$ $$\omega (X)=g(X,U).$$ Suppose $g(t)=g+ts$ and $\omega (t)=\omega +t\delta$, I want to linearize $|\pi |^2$. ‎By means of a local coordinate system on $M$ with local frame $\partial_i$‎, ‎set $‎\pi ‎(\partial_i,\partial_j)=‎\pi‎_{ij}$‎. ‎consequently‎, ‎$$ ‎\pi_{ij}=‎\nabla_i ‎\omega‎_j ‎-\omega_i ‎\omega_j ‎+‎\dfrac{1}{2}g_‎{‎ij} ‎|\omega|^2‎ ‎$$ ‎where,‎‎‎ $‎\omega‎(\partial_i)=\omega_i$. ‎We ‎have‎ $$‎|\pi |^2=g^{ij}\pi_‎{‎ij}=\mathrm{div}(\omega) ‎‎+‎(‎‎\dfrac{n}{2}-1)|\omega |^2 ‎‎.‎$$‎‎ Consequently,‎ ‎$$ (|\pi (t) ‎‎|^2 )'(0)=‎<‎‎‎d ‎\omega +(‎‎\dfrac{n}{2}-1)\omega\otimes \omega‎,‎-‎s>+<g ,d\delta +(n-2)\omega\otimes \delta>.‎ ‎$$

Is my result correct?