TITLE: If $3x^2$ is the derivative of $x^3$, how can $f'(x)$ be a linear map?
QUESTION [5 upvotes]: Suppose I have the function $f(x)=x^3$. The derivative is obviously $f'(x) = 3x^2$. But $3x^2$ is nonlinear since 
$$f'(3x) = 27x^2$$
$$3f'(x)  = 9x^2$$
Therefore this isn't a linear map. 
Rudin defines the following 
$$f(x+h)-f(x) = f'(x)h + r(h)$$
where $r(h)$ is very small, and he says we can regard the derivative of $f$ at $x$ as the linear operator that maps 
$$h\mapsto f'(x)h$$     
How does this make sense?

REPLY [7 votes]: $h\mapsto f'(x)h$ is a linear map of $\boldsymbol{h}$, not of $x$, which is only a fixed parameter in such questions. Isn't the tangent a straight line anyway?

REPLY [2 votes]: It is linear at each point.  For example $f'(2)=12$.  So, the linear map this induces is $x\mapsto 12x$.