Proof that $U(n)$ is a manifold

Question

I am trying to follow a proof that $U(n)$ is a Lie group. I have some trouble understanding some part of the proof that $U(n)$ is a manifold.

Define $F:M_n(\mathbb{C})\to H_n(\mathbb{C})$ by $F(X)=X^{*}X$, where $H_n(\mathbb{C})$ denotes the Hermitian $n\times n$ matrices over $\mathbb{C}$.

Then $\dim H_n(\mathbb{C})=n+\frac{n(n-1)}{2}=n^2$. This I understand.

In order to use the regular value theorem, we show the identity matrix $I_n$ is a regular value of $F$.

I recall from when I learned multivariable calculus that the differential of $F$ at $X$ is given by $$d_XF(A)=\lim_{A\to0}\frac{||F(X+A)-F(X)-\phi A||}{||A||}$$ where $\phi$ is a bounded linear operator.

But in the lecture notes I am reading, the author has written $$\dfrac{d}{dt}F(X+tA)\mid_{t=0}=\ldots$$

I follow his calculation of $\dfrac{d}{dt}F(X+tA)\mid_{t=0}$ and the remainder of the proof, but I do not understand why he has calculated this.

Question: Why calculate $\dfrac{d}{dt}F(X+tA)\mid_{t=0}$ instead of $d_XF(A)$?

Answer 1

It's the same thing, since $d_XF(A)=\frac{\mathrm d}{\mathrm dt}F(X+tA)|_{t=0}$.