Difference between gradient and derivative.

Question

My question may be a bit stupid, but this morning I tried to explain the gradient to someone, and he makes a parallel with derivative of function $f:\mathbb R\to \mathbb R$. What he says is that for a function $f:\mathbb R\to \mathbb R$, the gradient and the derivative are the same. I agree that the scalar value are the same, but I'mnot sure that the meaning behind is the same.

For example, take $f(x)=x^2$. For me the gradient of $f$ is going to be the vector field $\nabla f(x)=2x\cdot 1$, where $1$ is the basis of $\mathbb R$, so it should look like that

whereas the derivative $f'(x)$ is really the rate of the function, and if it would be a vecteur field, it would be a vector field over the range of $f$, and not on the domain of $f$ as the gradient is. What do you think ?

To illustrate, I would say that the derivative field is in red and blue, and the gradient is in pink.

Answer 1

Conceptually, you can define the derivative of a function $f: V \rightarrow W$ at a point $p\in V$ (where $V$ and $W$ are Banach vector spaces) as a linear operator $f'(p) \in L(V,W)$ such that: $$ \lim_{h\rightarrow 0} \frac{||f(p+h)-f(p)-f'(p)\cdot h||_W}{||h||_V} = 0 $$ You can then define the derivative of $f$ (without specifying a point) as a function $f' : V \rightarrow L(V,W)$.

The gradient can be defined for a function $f: M\rightarrow \mathbb R$, where $M$ is a Riemannian manifold with metric $g$. The gradient of a function $f$ at a point $p\in M$ is a vector $\nabla f(p) \in T_{p}M$ such that for any curve $\gamma :\mathbb R\ni t \mapsto \gamma(t) \in M$ with $\gamma(0) = p$ you have $$ g\Big(\nabla f(p), \frac{d\gamma}{dt}(0)\Big) = (f\circ\gamma)'(0)$$ where $\frac{d\gamma}{dt}$ is a vector tangent to the curve $\gamma$. Therefore the gradient (without specifying a point) is a vector field $\nabla f: M\rightarrow TM$.

If $V=M=\mathbb R^n$ and $W=\mathbb R$, then $T_pM \cong L(V,W)$ and both the derivative and the gradient can be defined and if they exist they will be equal to each other.

In the even more special case where $V=\mathbb R$ at every point $p$, you can furthermore identify $T_pM \cong \mathbb R$ and define the function $f' : \mathbb R\rightarrow \mathbb R$. However, you'd still traditionally see the gradient as a vector-valued function $\nabla f: \mathbb R \rightarrow T\mathbb R$. So while I agree that the pink/purple arrows are an appropriate representation of the gradient of $f(x)=x^2$, I would represent the derivative as a graph of the function $f'(x) =2 x$.