On a step of a proof of the mean value theorem in several variables. I am following this proof of the mean value theorem

And I am a bit stuck on the last line, to be precise, why is $$f(a +t[b-a] + h[b-a]) - f(a + t[b-a]) = \bigtriangledown f(a+t[b-a]) (h[b-a]) + o(h [b-a]) ?$$
I am familiar with little oh notation but I need a reminder on what happened here, does someone mind to lend a hand?
 A: The function $f$ is supposed differentiable at each point of the line segment $ab$, so, for every $x \in ab$, there exists a (necessarily unique) vector denoted by $\nabla f(x)$, such that $$f(x+v)-f(x)=\nabla f(x) \cdot v+o(v)$$ where $v$ is a vector.
The meaning of $o(v)$ is $$\lim_{\|v\| \rightarrow 0} \frac {f(x+v)-f(x)-\nabla f(x) \cdot v}{\|v\|}=0$$
In order to show that $g$ is differentiable at $t \in ]0,1[$, one needs to prove the existence of a (necessarily unique) number, denoted by $g'(t)$, such that $$g(t+h)-g(t)=g'(t) h+o(h)$$ where $h$ is a number.
The meaning of $o(h)$ is $$\lim_{h \rightarrow 0} \frac {g(t+h)-g(t)-g'(t)h}{h}=0$$
We want to prove that $g'(t)$ exists and $$g'(t)=\nabla f(a+t(b-a)) \cdot (b-a)$$ One has $$g(t+h)-g(t)=f(a+t(b-a)+h(b-a))-f(a+t(b-a))$$ Note that here $a+t(b-a)$ and $h(b-a)$ work as $x$ and $v$ respectively so $$g(t+h)-g(t)=\nabla f(a+t(b-a)) \cdot h(b-a)+o(h(b-a))$$ but a property of the dot product gives $$\nabla f(a+t(b-a)) \cdot h(b-a)=(\nabla f(a+t(b-a)) \cdot (b-a))h$$
Moreover the vector $b-a$ is constant so $$o(h(b-a))=o(h)$$ because banally $$\lim_{h \rightarrow 0}\, \frac {\varphi (h)}{h \|b-a\|}=0$$ iff $$\lim_{h \rightarrow 0} \frac {\varphi (h)}h=0$$ Then $$g(t+h)-g(t)=(\nabla f(a+t(b-a)) \cdot (b-a))h+o(h)$$
A: In n dimensions, with f mapping a subset of $R^n$ into $R$, the definition of $f$ being differentiable on the open  line segment joining $a$ to $b$ is equivalent to the last equation, whenever  $0<t<1$ and $0<t+h<1$, where $\nabla f(v)$ is a linear operator  from $R^n$ to $R$. Consider $t$ to be a fixed value in the last line, as we are trying to show that the function $g(x)$ is differentiable at $x=t$.
