Problem with a directional derivative calculation

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The discussion revolves around the calculation of directional derivatives using two methods: the limit definition and the dot product of the gradient and direction vector. Participants noted that different results were obtained from these approaches due to the lack of continuity in the partial derivatives at the point (0,0). It was clarified that the dot product method is valid only for differentiable functions, which require continuous partial derivatives. Since the function in question is not differentiable at (0,0), the limit method should be used instead. Understanding these concepts is crucial for accurate directional derivative calculations.
Amaelle
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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)

1605103234654.png


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!
 
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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.
 
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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?
 
Amaelle said:
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: $$g'(t) = \cos \theta \ \sin^2 \theta$$
 
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Likes Amaelle
yes thank you, understanding things takes times :) but thank you for easing the pain
 
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Amaelle said:
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.
 
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thanks a million!
 
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