Check that $df_x(v) = (v,v).$

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Here is a proof that I am totally different from my classmates'. So I am requesting for expert reference here. Thank you. :-)

Let $f: X \rightarrow X \times X$ be the mapping $f(x) = (x,x).$ Check that $df_x(v) = (v,v).$

$\begin{eqnarray*} df_x(v) &=& \lim_{t \rightarrow 0} \frac{f(x + tv) - f(x)}{t}\\ & = & \lim_{t \rightarrow 0} \frac{(x+tv, x+tv) - (x,x)}{t}\\ & = & \lim_{t \rightarrow 0} \frac{(tv,tv)}{t}\\ & = & \lim_{t \rightarrow 0} (v,v)\\ & = & (v,v). \end{eqnarray*}$

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If you replace $x + tv$ by a curve through $x$, $\gamma$ such that $\gamma(0)=x$ and $\gamma '(0)=v$, you would get the standard proof of the result.

In an arbitrary manifold $X$, that is the only way to make sense of the expression $x+tv$.