TITLE: Doubt in understanding Differentials
QUESTION [5 upvotes]: I am studying smooth manifolds, and recently I got to know about tangent spaces and the differentials.
Suppose $f:M \to N$ is a smooth map, then it induces a differential between the tangent spaces $df:T_cM \to T_{f(c)}N$ which is a linear transformation. This is what I read.
Now I was trying some examples, and I am stuck here. If I consider $f:\mathbb{R} \to \mathbb{R}$ defined by $f(c)=3c^3$, then the differential map at $x=c$ is $f'(c)=9c^2$. Hence the constant map $9c^2:\mathbb{R} \to \mathbb{R}$ is a linear map, which is not the case. I cannot figure out where I am going wrong.

REPLY [3 votes]: There are two possible misconceptions here. The first is that the map defined as $T_pf:T_pM\to T_{f(p)}N$, sometimes called differential, in reality is a different map than the differential $df$ of a function (this is way I had used two different notations). The first map is also called the tangential map of $f$ at $p$, and it is a linear map that maps vectors from a tangent space to other tangent space.
However the differential is defined by $d_pf v:=\pi_2\circ  (T_pf(p,v))$ where $\pi_2$ is the projection of the second argument and $(p,v)\in T_pM$, that is, we have that $T_pf(p,v)=(f(p),d_pf v)$. The maps are very close one to each other but they aren't exactly the same, this is way they generally are both called differential.
Now: the differential at $p$ is a function, if $f:\mathbb{R}^n\to \mathbb{R}^m$ then $d_pf$ is a linear map that can be represented by a Jacobian matrix (the matrix that represents the derivative of $f$ at $p$). If $n=m=1$ then the derivative is just a real number and so the function is just multiplication by an scalar, that is, $d_pf$ will be the map $x\mapsto f'(p)x$.