Why the derivative of $T(\gamma(s))$ is $T$ if this composition is not a linear transformation?

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Let $T:\mathbb{R}^3\rightarrow\mathbb{R}^3$ be a linear transformation and $\gamma:I\subset\mathbb{R}\rightarrow\mathbb{R}^3$ a p.b.a.l. curve.

I'm always needing to derivate the composed function $T(\gamma(s))$, so by the chain rule we get $d(T(\gamma(s)))=dT(\gamma(s).d(\gamma(s))$ and then my professor conclude that as $T$ is linear we have $dT(\gamma(s)=T$ and then

$d(T(\gamma(s)))=T\gamma^{'}(s)$ by i dont understand why because $d(T(\gamma(s)))$ is not even linear since $\gamma$ is not