Show that the tangent space of the diagonal is the diagonal of the product of tangent space

I'm stuck on this question for quite a few days and still haven't got a clue what to do. The question is as follows:

If $\Delta$ is the diagonal of $X\times X$ where $X$ is a manifold, show that its tangent space $T_{(x,x)}(\Delta)$ is the diagonal of $T_x(X)\times T_x(X)$.

Because this question follows a previous part, so I constructed a map

$$f:X\longrightarrow X\times X$$ such that f(x)=(x,x). Therefore I have

$X\overset{f}{\longrightarrow} X\times X$.

Then we take the derivative map

$$T_x(X)\overset{df_x}{\longrightarrow} T_{(x,x)}(X,X)$$

However this does not give me the tangent space of the diagonal...

• Your map $df_x$ is probably not surjective. What could be its range? – Thibaut Dumont Jun 7 '13 at 14:09
• Well, I know $f$ is probably not surjective. But I don't know what would happen when taking $df_x$. The diagonal is a manifold itself. So it should have a tangent space at (x,x) as well. Isn't it weird that two manifolds with different dimensions have the same tangent space? Thanks a lot for your help! – Evariste Jun 7 '13 at 14:11

Choose $(v,v) \in \Delta(T_x(X) \times T_x(X)$, with $v \in T_x(X)$. Then there is a smooth curve $\gamma : I \to X$ such that $\gamma(0) = x$, $\gamma'(0) = v$. Now $(\gamma(t) , \gamma(t))$ is a smooth curve that lives in the diagonal of $X \times X$. What happens when you differentiate the curve at $t = 0$? Prove the other direction in the same way.