# Difference between gradient and derivative.

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.

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 $$f'(x) =2 x$$.