Why is the component of a differentiable function also differentiable? I'm self studying with the Spivak book, and I'm trying to see a result that I know to be true but I'm having some time seeing how the maths pans out.  Supposing a function $f$ is differentiable.  Then given the projection operator $\pi^i$ where $\pi^i(f)=f^i$, $f^i$ must also be differentiable.
I've been trying to work it out by applying the chain rule
$$Df^i(a)=D(\pi^i\circ f)(a)=D\pi^i(f(a))\circ Df(a)$$
since the projection operator is linear, it's derivative is simply itself
$$Df^i(a)=\pi^i(f(a))\circ Df(a)$$
$$Df^i(a)=f^i(a)\circ Df(a)$$
I'm fairly sure the steps leading up to this point are correct, however it doesn't seem to make sense.  For example, let $f:\mathbb{R}\rightarrow\mathbb{R}^2$
$$f=\begin{bmatrix}x^2\\x\end{bmatrix}$$
then
$$Df=\begin{bmatrix}2x\\1\end{bmatrix}$$
So if we let $i=1$,
$$\begin{align}
Df^1(a)&=f^1(a)\circ Df(a) \\
&=a^2\circ\begin{bmatrix}2a\\1\end{bmatrix}
\end{align}$$
Which I am certain is wrong, since $Df:\mathbb{R}\rightarrow\mathbb{R}^2$, but $f^1:\mathbb{R}\rightarrow\mathbb{R}$ so it doesn't match up.  I think I have misunderstood either the derivative of the projection operator or how $f^i(a)$ forms a composition with $Df(a)$.  Could someone point me in the right direction?  Thanks!
 A: You're misusing some notation, and probably misunderstanding what it means to say "derivative of a linear transformtion at every point is itself". In the beginning you write $\pi^i(f)=f^i$. This is not at all what Spivak writes. $\pi^i$ is a function $\Bbb{R}^n\to\Bbb{R}$, so its input is a tuple of numbers $(a^1,\dots, a^n)$ and its output is $a^i$. By definition if we have a function $f:\Bbb{R}^k\to\Bbb{R}^n$, then by $f^i$, we mean the function $\pi^i\circ f$. Now,
\begin{align}
(Df^i)(a)&=D(\pi^i\circ f)(a)=(D\pi^i)(f(a))\circ Df(a)=\pi^i\circ Df(a),
\end{align}
where the last equality is because $\pi^i:\Bbb{R}^n\to\Bbb{R}$ is a linear transformation, so at any point (for example $f(a)$) its derivative is itself. This equality makes sense, because $f^i:\Bbb{R}^k\to\Bbb{R}$, so $Df^i(a):\Bbb{R}^k\to\Bbb{R}$ is linear. On the RHS we have a composition $Df(a):\Bbb{R}^k\to\Bbb{R}^n$ with $\pi^i:\Bbb{R}^n\to\Bbb{R}$.
For your example below, Spivak uses the $D$ notation to mean the linear transformation and the prime notation to refer to the corresponding matrix representations relative to the standard bases. So, if $f:\Bbb{R}\to\Bbb{R}^2$ is given as $f(x)=(x^2,x)$, then $Df(a):\Bbb{R}\to\Bbb{R}^2$ is the linear transformation
\begin{align}
Df(x)[h]&=(2xh,h)
\end{align}
And so $Df^1(a)[h]=(\pi^1\circ Df(a))[h]=\pi^1(2ah,h)=2ah$.
Or if you prefer matrices, then
\begin{align}
(f^1)'(a)&=(\pi^1)'(f(a))\cdot f'(a)\\
&=[\pi^1]_{\text{standard bases}}\cdot f'(a)\\
&=
\begin{pmatrix}
1&0
\end{pmatrix}
\cdot
\begin{pmatrix}
2a\\
1
\end{pmatrix}\\
&=2a,
\end{align}
where $[\pi^1]$ is the matrix representation of the linear map $\pi^1$ with respect to the standard basis $\{(1,0),(0,1)\}$ on the domain $\Bbb{R}^2$ and $\{1\}$ on the target space $\Bbb{R}$. This should of course make sense because $f^1(x)=(\pi^1\circ f)(x)=\pi^1(x^2,x)=x^2$, so the derivative at a point $a$ is $2a$.
