# Homework Help: Directional derivative at a point

1. Oct 30, 2015

### catch22

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
part a) finding partial derivatives:

and plugging in (2,0,1) into each, I get the gradient which is <0,-2,0>

to find the directional derivative, it is the dot product of the gradient and unit vector of (3,1,1):

part b) isn't the direction of the greatest increase just the gradient, which I found in part A?

and the greatest increase at (2,0,1) is the magnitude of the gradient there, which is

2. Oct 30, 2015

### andrewkirk

Yes.
A pedant might insist that a direction should be given as a normalized vector. But we are not pedants.

3. Oct 30, 2015

### Ray Vickson

Some of your partial derivatives are incorrect (although not by very much).
One of your partial derivatives is wrong.

4. Oct 30, 2015

### geoffrey159

Also, you don't say anything about the continuity of the first partial derivatives in a neighborhood of $(2,0,1)$. It is under that condition that the formula $(D_{\vec v}f)(2,0,1) = (\nabla f. \vec v) (2,0,1)$ is valid.

5. Oct 30, 2015

### catch22

wow, I can't believe I didn't spot that +x at the end.

it should be (-yz e^-xyz) + 1

6. Oct 30, 2015

### catch22

the correct answers after fixing the partial derivative with respect to x

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7. Oct 30, 2015

### catch22

do you mean like saying the domain is from negative infinity to positive infinity?

8. Oct 30, 2015

### geoffrey159

Not at all and I wonder why you ask that ?

9. Oct 30, 2015

### Ray Vickson

Well, it IS possible to have a well-defined directional derivatives in some directions $\vec{d}$ at a point $\vec{r_0}= (x_0,y_0,z_0)$, but to not have well-defined, finite partial derivatives at $\vec{r_0}$. However, the present example is not so pathological.