how to prove that $f(p(x))=p'(x)$ is a linear transformation 
Let $f: \mathbb{R}_{n}[x] \mapsto \mathbb{R}_{n}[x]$ be an application define by $f(p(x))=p'(x)$. How to prove that is a linear transformation?

I did this way:
In order to $f \space$ be a linear transformation, $2$ conditions have to be verified.
$f(p(x)+q(x))=f(p(x))+f(q(x)) \space $ and $\space \lambda \space f(p(x))=f(\lambda(p(x))) \space \forall \lambda \in \mathbb{R}$
For the first:
$f(p(x)+q(x))=f((p+q)(x))=(p+q)'(x)=p'(x)+q'(x)=f(p)+f(q)$
For the second:
$ \lambda \space f(p(x))=\lambda p'(x) =f(\lambda p(x))$
I'm no sure if the second is rigth.Can you help me?Thanks 

EDIT:
I found this explanation for the first condition:
Let $p_{n}$ a polynomial whose degree do not exceed $n$. In its canonic form $p_{n}(x)=a_{0}x^n+a_{1}x^{n-1}+...+a_{n-2}x^2+a_{n-1}x+a_{n}$.
Then $ \space p_{n}'(x)=a_{0}nx^{n-1}+a_{1}(n-1)x^{n-2}+...+a_{n-2}2x+a_{n-1}$. In other words, $p'_{n}(x)$ is a polynomial whose degree do not exceed $n-1$.
$$f:\mathbb{R}_{n}[x] \mapsto \mathbb{R}_{n-1}[x]$$
Now let $q_{n}(x)$ and $p_{n}(x)$ be $2$ vectors in the same space. If the degree of $q(x)$ is not equal to the degree of $p(x)$ is because the coefficients of the inexistent degrees are zero. So,
$f(p(x))+f(q(x))=a_{0}nx^{n-1}+a_{1}(n-1)x^{n-2}+...+a_{n-2}2x+a_{n-1}+b_{0}nx^{n-1}+b_{1}(n-1)x^{n-2}+...+b_{n-2}2x+b_{n-1}$
By adding like terms,
$=(a_{0}+b_{0})nx^{n-1}+(a_{1}+b_{1})(n-1)x^{n-2}+...+(a_{n-2}+b_{n-2})2x+a_{n-1}+b_{n-1}$
That's equal to $f(p(x)+q(x))$
My doubt in the second condition is:
If ones multiplied each term of $p(x)$ by a real number $\lambda$, ones get a new coefficient for each term that is equal to $\lambda a_{k}$. Where $k=0,1,2,...,n$. These new coefficients will have the same behaviour of the previous. So,
$\lambda f(p(x))=\lambda p_{n}'(x)=\lambda a_{0}nx^{n-1}+\lambda a_{1}(n-1)x^{n-2}+...+\lambda a_{n-2}2x+\lambda a_{n-1}$
How can I integrate the $\lambda$ inside $f$?
 A: The application in question is the usual differentiation (in this case, it seems, polynomials). What you want to prove is that 
$$D(f+g)=D(f)+D(g)$$
and that
$$D(\lambda f)=\lambda D(f)$$
Note that for any polynomials $p(x)=\sum_{k=0}^n a_k x^k$ and $q(x)=\sum_{k=0}^n b_k x^k$ we have that
$$D(p(x)+q(x))=\sum (a_k+b_k) kx^{k-1}=\sum a_k kx^{k-1}+\sum b_k kx^{k-1}=D(p(x))+D(q(x))$$
For the other
$$D(\lambda p(x))=\sum \lambda a_k  kx^{k-1}=\lambda \sum a_k  kx^{k-1}=\lambda D(p(x))$$
Interpret both sum, multiplication by scalars and differentiation as an operation on the coefficients, and you'll get this much easier. We can encode a polynomial by its coefficients, say $p=(a_1,\dots,a_n),q=(b_1,\dots,b_n)$. Then we have the sum
$$p+q:=(a_0+b_0,\dots,a_n+b_n)$$
the multiplication by scalars
$$\lambda\cdot  p:=(\lambda a_0 ,\dots, \lambda a_n )$$
and differentiation
$$p':=(a_1 ,2a_2,\dots,(n-1)a_{n-1}, 0 )$$
You can try and see that with this idea (which is basically an isomorphism using coordintes over $\Bbb R^n$) we can prove all what you want, namely that
$$(p+q)'=p'+q'$$ and that $$(\lambda\cdot  p)'=\lambda\cdot  p'$$
