# Existence of directional derivative

• songoku
songoku
Homework Statement
Relevant Equations
Partial derivative

Direction derivative in the direction of unit vector u = <a, b>:
Du f(x,y) = fx (x,y) a + fy (x,y) b

My attempt:
I have proved (i), it is continuous since ##\lim_{(x,y)\rightarrow (0,0)}=f(0,0)##

I also have shown the partial derivative exists for (ii), where ##f_x=0## and ##f_y=0##

I have a problem with the directional derivative. Taking u = <a, b> , I got:
$$Du =\frac{\sqrt[3] y}{3 \sqrt[3] {x^2}}a+\frac{\sqrt[3] x}{3 \sqrt[3] {y^2}}b$$

Then how to check whether the directional derivative exists or not?

Thanks

The directional derivative at ##[0,0]## in direction ##[a,b]## (has to be zero since both partial directions are and) is given by
$$D_{[0,0]}(f)\cdot [a,b]=0=\lim_{h\to 0}\dfrac{f([0,0]+h[a,b])-f([0,0])}{h}$$
What do you get on the right-hand side for ##f([x,y])= \sqrt[3]{xy}##?

songoku
fresh_42 said:
The directional derivative at ##[0,0]## in direction ##[a,b]## (has to be zero since both partial directions are and) is given by
$$D_{[0,0]}(f)\cdot [a,b]=0=\lim_{h\to 0}\dfrac{f([0,0]+h[a,b])-f([0,0])}{h}$$
What do you get on the right-hand side for ##f([x,y])= \sqrt[3]{xy}##?
$$0=\lim_{h\to 0}\frac{\sqrt[3]{(h^{2}ab)}-0}{h}$$
$$0=\lim_{h\to 0}\frac{(ab)^{\frac{1}{3}}}{h^{\frac{1}{3}}}$$

The limit can only exist if ##(ab)^{\frac{1}{3}}=0## so either ##a=0## or ##b=0## and since ##u## is unit vector, if ##a=0## then ##b=\pm 1## and vice versa

Is my working correct? Thanks

fresh_42
Yes, that is correct.

songoku
Thank you very much fresh_42

jim mcnamara

## What is a directional derivative?

A directional derivative represents the rate at which a function changes at a point in the direction of a given vector. It generalizes the concept of a partial derivative by considering changes not just along the coordinate axes but in any specified direction.

## How do you compute the directional derivative of a function?

To compute the directional derivative of a function $$f$$ at a point $$\mathbf{a}$$ in the direction of a unit vector $$\mathbf{u}$$, you take the dot product of the gradient of $$f$$ at $$\mathbf{a}$$ with the vector $$\mathbf{u}$$. Mathematically, it is given by $$D_{\mathbf{u}} f(\mathbf{a}) = \nabla f(\mathbf{a}) \cdot \mathbf{u}$$.

## What conditions must a function satisfy for the directional derivative to exist?

For the directional derivative of a function $$f$$ at a point $$\mathbf{a}$$ in the direction of a vector $$\mathbf{u}$$ to exist, the function must be differentiable at $$\mathbf{a}$$. Differentiability implies that the gradient $$\nabla f$$ exists at $$\mathbf{a}$$ and the directional derivative can be computed.

## Can the directional derivative exist if the function is not differentiable?

Yes, the directional derivative can exist even if the function is not differentiable. However, in such cases, the existence of the directional derivative does not guarantee the existence of a well-defined gradient. The function may have directional derivatives in some directions but not be differentiable in the general sense.

## How is the concept of a directional derivative used in optimization problems?

In optimization problems, the directional derivative is used to determine the rate of change of the objective function in different directions. This information is critical in gradient-based optimization methods, such as gradient descent, where the gradient (composed of directional derivatives) guides the search for the minimum or maximum of the function.

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