Find the image of the vector v under the mapping f This question comes from Ordinary Differential Equations, Arnold, P67, and it asks
"Suppose v is a positive unit vector of the line attached at the point a and let $f(x) = x^2$. Find $f_{*a}v$."
I tried the following:
\begin{align*}
f_{*a}v &= \frac{d}{dt} \Big|_{t = 0} f(\phi(t)) \\
&= \frac{df(\phi(t))}{dt} \Big|_{t = 0} \cdot \frac{\phi(t)}{dt} \Big|_{t = 0} \\
&= 2a \Big|_{t = 0} \cdot v \\
&= 2at
\end{align*}
It is indeed the final answer but I don't know how 
$\frac{df(\phi(t))}{dt} \Big|_{t = 0} = 2a \Big|_{t = 0}$
or I did it wrong.
Here is the definition in the book:
The Image of the vector v under the mapping f is the velocity vector with which the moving point $f(\phi(t))$ leaves the point $f(x)$ when the moving point $\phi(t)$ leaves the point $x$ with velocity $v$:
$$
f_{*x} v = \frac{d}{dt} \Big|_{t = 0} f(\phi(t)), \quad \text{where } \phi(0) = x, \quad \frac{d}{dt} \Big|_{t = 0} \phi(t) = v
$$
Appreciate any help.
 A: See, the idea is that $f_{*a}$ is a linear transformation for each point $a$ on the curve. How does this work?
Recall that in multivariable calculus, the derivative of a function is a linear transformation : it takes a vector and returns the "infinitesimal change" of the function along that direction.
The better way of understanding things is that the derivative is the best linear approximation to a function at a point. So, if you have a function and a point, and you want to understand how the function behaves in a neighbourhood of this point, then you can use a "best" linear approximation of the function, which comes as a linear transformation.
For example, take $f(x) = x^2$. Here, the derivative is a "number", which does not sound like a linear transformation : but it is, for every number $a$ may be interpreted as the linear transformation $v \to av$(i.e. scaling) for any vector $v$.

Now, what is $\phi$? Note that $\phi$ is called a "moving point", but in truth it is a "parametrization". What does that mean?
Think of it this way : imagine you have a real number $x'$ close to $x$, and you want to find the value of the function at $x'$. For this, it would be nice if you had a "path" from $x$ to $x'$, which you could travel along .Think of $\phi$ as a path in that case, which starts from $x$ and goes to a point where you want to find the function value.
Now, that is why $\phi(0) = x$ : the path should start at $x$. But then, we also know that the "velocity" $v$ of the path is given : that is why we require $\frac{d \phi}{dt}\Big|_{t=0} = v$.
So what does this achieve? Well, this allows approximation when $x'$ lies in the direction $v$ from $x$. So, in  some sense, $\phi$ is the straight line starting from $x$ in the direction $v$! And the points we would like to approximate are points lying close to $x$ in this direction from $x$.

With that, the model requires $\phi(t) = x+tv$ for $t \in [0,\epsilon)$ with $\epsilon$ as small as required(It doesn't matter, we are taking the derivative at zero anyway) so that we have a straight line. 

Now, the best linear approxmation of $f$ is to be rephrased in terms of the parametrizing variable $t$ as $\frac{d(\phi(t))}{dt}|_{t=0}$, where $t = 0$ is because $\phi(0) = x$. Now, using the chain rule,
$$
\frac{d f(\phi(t))}{dt} = \frac{d(f(\phi(t)))}{d\phi(t)} \frac{d\phi(t)}{dt}
$$
And therefore, now we use the formula which tells us that the first fraction is just the derivative of $x^2$ at $a$ which is $2a$. The second fraction is the derivative of $x + tv$ w.r.t $t$ at $0$, which is $v$. Now, multiplying gives us $2av$, as desired.
More precisely, what does this mean? This means, that $f(x') \approx f(x) + 2a\bar{v}|v|$, where $\bar v$ is the unit vector in the direction $x'-x$ is, and $|v|$ is the magnitude of $x'-x$.
Note that in our case, since $x,x'$ are all real numbers, $|v|$ is just $|x'-x|$, and $\bar v$ is $\pm 1$ depending upon whether $x'$ is to the left or the right of $x$.
