For a function $f:\Bbb R^n→\Bbb R$, Differential $f(x)$ is a linear function. But why?

Question

I just read that for a function $f: \Bbb R^n → \Bbb R$, the differential $df_x$ is a linear function for every $x\in\Bbb R^n$. I can't convince myself why that's the case. In particular, I don't understand what $\Bbb R^n → \Bbb R$ has to do with it.

Answer 1

For a function $f:\Bbb R^n\to\Bbb R^m$, $df_x$ is by definition the linear map $\Bbb R^n\to \Bbb R^m$ such that $\lim_{h\to 0}\frac{\lVert f(x+h)-f(x)-df_x(h)\rVert}{\lVert h\rVert}=0$. This does not absolutely mean that, for a differentiable function $f:\Bbb R^n\to\Bbb R^m$, the map $$d f_\bullet:\Bbb R^n\to\{\text{linear maps }\Bbb R^n\to\Bbb R^m\}\\ x\mapsto df_x$$ is linear. In fact, it hardly ever is: in short, you shouldn't expect $df_{\lambda x+\mu y}=\lambda df_x+\mu df_y$.

However, for each $x$ the map $$\{\text{differentiable functions at }x\}\to\{\text{linear functions }\Bbb R^n\to\Bbb R^m\}\\ f\mapsto df_x$$ is linear: in other words, $d(\lambda f+\mu g)_x=\lambda df_x+\mu dg_x$.

Answer 2

That's the definition of the differential, so I'm unsure why any convincing is needed. The differential $\mathrm d_xf$ of a function $f:\mathbb R^n\to\mathbb R$ is defined as the unique linear map $\mathrm d_xf:\mathbb R^n\to\mathbb R$ such that

$$\lim_{h\to0}\frac{f(x+h)-f(x)-\mathrm d_xf(h)}{\Vert h\Vert}=0.$$

The reason why it makes sense to define it that way is that the differential is the generalization of the derivative in one dimension, which is essentially a linear approximation of the function: The function is approximately $f(x+h)\approx f(x)+f'(x)\cdot h$. And a linear approximation of a higher dimensional function is then $f(x+h)\approx f(x)+\mathrm d_xf(h)$. The above equation is just a formal definition of what $\approx$ means.

Answer 3

I don't understand what you mean by "What does $R^n\to R$ have to do with it? "! That is telling you what the setting of the question is! If n= 1 we have the standard "Calculus 1" problem of differentiating a function of single real number to the real numbers such as "$f(x)= e^x$". Its derivatve is, of course, $f'(x)= e^x$ and its "differential" is $df= e^x dx$. At a specific x= a, $df= e^a dx$, which we can think of as the linear operation "multiply dx by the number, $e^a$. If n= 2, f is a real valued function of two variables such as $f(x,y)= x^2+ y^2$. Its "derivative" is the gradient, $\nabla f= 2x\vec{i}+ 2y\vec{j}$ and it "differential" is $2x dx+ 2y dy= (2x\vec{i}+ 2y\vec{j})\cdot (dx\vec{i}+ dy\vec{j})$. That is, the differehtial is the linear operation of "multiplication by $2x\vec{i}+ 2y\vec{j}$.