I'm trying to show that the map $f: S^n \to \mathbb{R}P^n$ given by sending a unit vector $x$ in $S^n \subset \mathbb{R}^{n+1}$ to the line spanned by $x$ in $\mathbb{R}P^n$ has injective derivative.
My attempt so far has been to let $i: S^n \to \mathbb{R}^{n+1}$ be the inclusion map and $\pi: \mathbb{R}^{n+1} \to \mathbb{R}P^n$ be the quotient map and then we can write $f = \pi \circ i$. Using this we have that the derivative is given by: $T_xf = T_x\pi \circ T_xi$. Now in order to show that $T_xf$ is an isomorphism it's sufficient to show it's injective (dimensions match up) so if $v$ is such that $T_xf(v) = 0$ then we have $T_xi(v)$ is in the kernel of $T_x\pi$ but do we know that $T_x\pi$ is injective?
Perhaps there is a simpler way to argue this but I haven't found it. Can we deduce directly that because $f$ is surjective then it's derivative is surjective?
We can continue on your way.
Though - as it is remarked, $T_x\pi$ is not injective - we can figure out its kernel:
By dimension considerations, it is one dimensional, and as moving align on the line of vector $x$ (and the origin) in $\Bbb R^{n+1}$, its image is staying in one point, hence (after identifying $T_x\Bbb R^{n+1}=\Bbb R^{n+1}$), $x$ itself will generate the kernel.
However, $x$ is orthogonal to the tangent hyperplane of the sphere $im\, T_xi$, in particular, $x\perp v$, which together proves that $v=0$.