Relationship between gradient and position vector

Question

When we take a derivative of the function f(x) with respect to x we find out how much an infinitely small change in x will change f(x).

When we take the gradient of a multivariable function say f(x,y,z) we define it as

$$ \nabla.f(x,y,z) = \frac{\partial f(x,y,z)}{\partial x} \hat{i} + \frac{\partial f(x,y,z)}{\partial y} \hat{j} + \frac{\partial f(x,y,z)}{\partial z} \hat{k} $$

and if the position vector is given by $ \vec{r}(x,y,z) = x\hat{i} + y\hat{j} + z\hat{k} $

Then is it safe to correlate that the gradient is the directional derivative of the function with respect to the position vector $\vec{r}$

If I am wrong can you explain in detail why ?

Answer 1

No, because the directional derivative of a function does not change it's "type", e.g. the directional derivative of a scalar-valued function is a scalar-valued function, not a vector-valued function.

But choose any vector $\vec v$ and calculate $v\cdot\nabla f$. You will find that this is the directional derivative (and perhaps you should review the definition of directional derivative). So the quantity you want is $\vec r\cdot\nabla f$; but note that $\vec r$ does not get differentiated in any way.

Answer 2

When $f$ is a function of a single variable, there is only one direction (up to sign) in which one can differentiate; therefore one typically speaks of differentiating $f$ with respect to $x$. But in several variables, the position vector $\vec r$ (i.e., the point at which one studies the change in the function) and the direction $v$ (in which the function is differentiated) are unrelated vectors and should not be confused. What is true is that the directional derivative in the direction $v$ is given in terms of the gradient $\nabla f$ at the point $\vec r$ by the formula $v \cdot \nabla f$.

Note that the gradient at the point cannot be "the directional derivative in the direction of the position vector", since the former (gradient) is a vector whereas the latter (directional derivative in the direction of the position vector) is a scalar.