For the following function: $f(x,y)=\ln(x-y)$, at $(2,1)$ what is the unit vector for the minimum directional derivative of $f$ ?

Firstly to find the gradient of $f$ at $(2,1)$: $$ \nabla f(x,y) = \begin{bmatrix} \frac{\partial}{\partial x}(\ln(x-y))\\ \frac{\partial}{\partial y}(\ln(x-y))\\ \end{bmatrix} = \begin{bmatrix} \frac{1}{x-y}\\ \frac{-1}{x-y}\\ \end{bmatrix} $$

$$ \nabla f(2,1) = \begin{bmatrix} 1\\ -1\\ \end{bmatrix} = \langle 1,-1\rangle $$

The minimum directional derivative is: $$-|\nabla f(2,1)| = -|\langle 1,-1\rangle|=-\sqrt 2$$

To find the unit vector: $$ u = \frac{\nabla f(2,1)}{-|\nabla f(2,1)|} = \frac{\langle 1,-1\rangle}{-\sqrt2} = \frac{\langle -1,1\rangle}{\sqrt2} $$

$\therefore$ The minimum directional derivative of $f(x,y)$ is $-\sqrt 2$ and occurs in the direction of unit vector $\frac{\langle -1,1\rangle}{\sqrt2}$

Is this the correct procedure to solve the problem? Thanks in advance.


1 Answer 1


Your proof is correct but is, in my opinion, missing a few words here and there explaining why your procedure is correct.

As an example : it seems to me that you're implicitly using the property that the directional derivative $\partial_v f$ at a point $a$ is the dot product $\nabla f (a) \cdot v$. This is not always the case ! But it is true when the function $f$ is differentiable at $a.$ So your proof should at least contain the words " Since $f$ is differentiable at ..."

Similarly you could be elaborate a bit more on certain parts:

The minimum directional derivative is: $$−|\nabla f(2,1)|=−|⟨1,−1⟩|=−\sqrt 2$$

Why is this true ? You don't need the write a huge paragraph of course ;) but a few extra words here and there would make your proof far more readable.


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