If $\beta: \mathbf{U}\times \mathbf{V}\to \mathbf{W}$ is a bilinear map between real linear spaces then its derivative at a point $(u,v)$ is given by the Leibniz rule $$D\beta(u,v)(u\_0,v\_0)=\beta(u,v\_0)+\beta(u\_0,v).$$


Consequently the derivative of the multiplication map of a real linear space $\mu:\mathbb R\times \mathbf{V}\to \mathbf{V}$ is dual number multiplication $D\mu (\lambda,x)(\alpha,v)=\lambda v+\alpha x$.


Now suppose $\pi:E\to Y$ is a vector bundle, with multiplication map $\mu:\mathbb R\times E\to E$ over $Y$. Is it possible to calculate the tangent map $\mathrm T \mu: \mathrm T(\mathbb R \times E)\to \mathrm TE$, even though the multiplication map is only fiberwise bilinear?


*Remark.* My question is motivated by [this one](https://mathoverflow.net/q/374823/69037), whose answer I do not understand.



The tangent map of the multiplication map of a vector bundle
Go back to the definition. Let $\gamma$ be a curve in $\mathbb{R}\times E$, we can take local coordinates on $E$ using the local trivialization, so
$$ \mathbb{R} \times E \ni (\lambda(s), v(s), b(s)) = \gamma(s) $$
The multiplication map gives
$$ \mu(\gamma(s)) = (\lambda(s) v(s), b(s)) $$
From this you find immediately that the derivative at $0$ is
$$ (\dot{\lambda}(0) v(0) + \lambda(0) \dot{v}(0), \dot{b}(0) ) \tag1$$




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More generally: the tangent space to a Cartesian product $A\times B$ at a point $(a,b)$ is isomorphic to $T\_a A \oplus T\_b B$.


Now, take $\varphi:A\times B \to C$. Fix $a\in A$ and $b\in B$. Let $\sigma(\bullet) = \varphi(\bullet, b)$ and $\tau(\bullet) = \varphi(a,\bullet)$. Then
$$ T\_{(a,b)}\varphi(\alpha,\beta) = T\_a\sigma(\alpha) + T\_b\tau(\beta) $$
*exactly like in multivariable calculus*.


In your case, with $\varphi$ the multiplication operator, you have that $\tau$ is the smooth map $E\to E$ given by multiplication by the scalar $a$. And $\sigma$ is the the map $\mathbb{R}\to E$ given by multiplying the fibre of a fixed element $b\in E$.


Perhaps what is confusing you is the fact that given a vector bundle $E \to Y$ and $e\in E$, that there is (via the linear structure on the fibre) a canonical identification of the fibre $\pi^{-1}(\pi(e))$ over $\pi(e)$ with the vertical part $V\_eE$ of the tangent space (the kernel of $T\_e\pi$ as a mapping from $T\_e E \to T\_{\pi(e)} Y$). If you denote by $\psi\_e: \pi^{-1}(\pi(e)) \to V\_e E$ this mapping, then you can write (1) as


$$ T\_{(\lambda,e)}\mu(\alpha,v) = \lambda v + \alpha \psi\_e(e) $$


