# Showing this function is differentiable in the origin

$$f:\mathbb{R^2}\to\mathbb{R}$$ a differentiable function in the origin so:

$$f(t,t) =t^3+t$$ and $$f(t,-2t)=2t$$

Calculate $$D_vf(0,0)$$

$$v=(1,3)$$

I have no idea on how to approach this problem. I know that because f is differentiable we have

$$D_vf(0,0)= Df(0,0)v$$

So I should be able to determine the partial derivatives. But how can I do it? I just need a hint to start the problem, thanks

• Does "Calculate $D_vf(0,0)v=(1,3)$" mean that we have to determinate $v=(v_1,v_2)$ such as the equation is satisfied ? – hamza boulahia Jan 28 '17 at 15:57
• Oh no, no! Sorry I didn't realize that bit wasn't clear when I was writing it. It just means that the vecto $v$ is $(1,3)$ – Granger Obliviate Jan 28 '17 at 16:19

## 1 Answer

I was looking for a function that verifies the given condition.

It is clear that this function is most likely a polynomial. So i assumed that, $$f(x,y)=ax^3+by^3+cx^2y+dxy^2+exy+kx+ly+m,\quad a,b,c,d,e,k,l\ \mbox{and}\ m\in\mathbb{R}$$ After verifying the conditions, i find the following function $$f(x,y)=\frac{2}{3}x^2y+\frac{1}{3}xy^2+\frac{4}{3}x+\frac{1}{3}y$$

Now we calculate, $\nabla f(x,y)$, $$\nabla f(x,y)=\left(\begin{array}[c] \ \frac{\partial f}{\partial x}\\ \frac{\partial f}{\partial y} \end{array}\right)=\left(\begin{array}[c] \ \frac{4}{3}xy+\frac{1}{3}y^2+\frac{4}{3}\\ \frac{2}{3}x^2+\frac{2}{3}xy-\frac{1}{3} \end{array}\right)$$ Then, $$\nabla f(0,0)=\left(\begin{array}[c] \ \frac{4}{3}\\ -\frac{1}{3} \end{array}\right)$$ And we have $D_vf(0,0)=\nabla f(0,0)\cdot v$, hence $$D_vf(0,0)=\left(\begin{array}[c] \quad \frac{4}{3}\\ -\frac{1}{3} \end{array}\right)\cdot \left(\begin{array}[c] \ 1\\ 3 \end{array}\right)=\frac{1}{3}$$

Remark:

The function $f$ that i found is not the only one that satisfies the conditions.