Problem with a directional derivative calculation

[Amaelle]

Homework Statement

calculate the directional derivative

Relevant Equations

directional derivative

Good day

I have a problem regarding the directional derivative (look at the example below)

in this example, we try to find the directional derivatives according to the two approaches ( the definition with the limit and the dot product of the vector gradient and the vector direction)
in this example, we got totally different results?

why?

many thanks in advance!

 

[pasmith]

The partial deivatives are not continuous: for $x= 0$ and $y \neq 0$ you have $\frac{\partial f}{\partial x} = \frac{y^4}{y^4} = 1$ which does not tend to 0 as $y \to 0$.

 

[Amaelle]

Thanks a lot for your prompt answer
so If I understand well, the dot product method works only if the function is differentiable? otherwise, we should only use the limit method?

 

[PeroK]

in this example, we try to find the directional derivatives according to the two approaches ( the definition with the limit and the dot product of the vector gradient and the vector direction)
in this example, we got totally different results?

why?

many thanks in advance!

What your book is doing is calculating the derivative along a parameterised curve through $(0,0)$:$$g(t) = f(t\cos \theta, t\sin \theta) = t\cos \theta \ \sin^2 \theta$$ Which is well-defined. And, trivially: $$g'(t) = \cos \theta \ \sin^2 \theta$$

 

[Amaelle]

yes thank you, understanding things takes times :) but thank you for easing the pain

 

[Delta2]

Thanks a lot for your prompt answer
so If I understand well, the dot product method works only if the function is differentiable? otherwise, we should only use the limit method?

Yes. A function of many variables is differentiable if and only if all the partial derivatives exist and are continuous . But in this case as post #2 shows the partial derivatives are not continuous at (0,0) hence the function is not differentiable at (0,0) hence you can't apply the gradient approach for calculating the directional derivative.

 

[Amaelle]

thanks a million!