Directional Derivatives — Question 7

Question 7

Define $$f(x,y)= \begin{cases} \dfrac{x^3}{x^2+y^2},&(x,y)\ne(0,0),\\ 0,&(x,y)=(0,0). \end{cases}$$ Tasks

1. Compute $D_{u}f(0,0)$ from the definition for every unit $u=\left\langle\cos\theta,\sin\theta\right\rangle$.

2. Compare with $\nabla f(0,0)\cdot u$.

3. Explain what this proves about directional derivatives and differentiability.

Question 7 – Solution

Strategy. Approach the origin along the ray $hu$ and keep the signed parameter $h$.

Step 1: Definition For $h\ne 0$, $$f(h\cos\theta,h\sin\theta) =\frac{h^3\cos^3\theta}{h^2} =h\cos^3\theta.$$ Therefore $$\boxed{D_{u}f(0,0)=\lim_{h\to 0}\frac{h\cos^3\theta}{h} =\cos^3\theta}.$$

Step 2: Gradient comparison Axis definitions give $f_x(0,0)=1$ and $f_y(0,0)=0$, so $$\nabla f(0,0)\cdot u=\cos\theta,$$ which differs from $\cos^3\theta$ for most angles.

Conclusion Every directional derivative exists, but its dependence on $u$ is not linear. Thus $f$ is not differentiable at the origin, and the gradient formula is not valid there.