# Thread: differentiability multivariable function

1. ## differentiability multivariable function

Hi, I need help on the following:

Let $\displaystyle f : \mathbb{R}^2 \to \mathbb{R}^2$ be defined by $\displaystyle f(x,y) = (y, x^2)$. Prove that it is differentiable at $\displaystyle (0,0)$.

I know the partial derivatives exist at $\displaystyle (0,0)$.

Thanks

2. Originally Posted by storchfire1X
Hi, I need help on the following:

Let $\displaystyle f : \mathbb{R}^2 \to \mathbb{R}^2$ be defined by $\displaystyle f(x,y) = (y, x^2)$. Prove that it is differentiable at $\displaystyle (0,0)$.

I know the partial derivatives exist at $\displaystyle (0,0)$.

Thanks
So the derivative is going to be a 2 by 2 matrix.

$\displaystyle \begin{bmatrix} 0 & 1 \\ 2x & 0 \end{bmatrix}$

If you evaluate this at zero we get

$\displaystyle m=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$

Now you need to calculate the limit

$\displaystyle f'(\mathbf{0})=\lim_{\mathbf{h} \to \mathbf{0}}\frac{||f(\mathbf{0}+\mathbf{h})-f(\mathbf{0})-m\mathbf{h}||}{||\mathbf{h}||}=0$

and show that it is equal to zero!

Where

$\displaystyle \mathbf{h}=\begin{bmatrix}h_1 \\ h_2 \end{bmatrix}$

This should get you started.

3. Originally Posted by storchfire1X
Hi, I need help on the following:

Let $\displaystyle f : \mathbb{R}^2 \to \mathbb{R}^2$ be defined by $\displaystyle f(x,y) = (y, x^2)$. Prove that it is differentiable at $\displaystyle (0,0)$.

I know the partial derivatives exist at $\displaystyle (0,0)$.

Thanks
You can also use the fact that if $\displaystyle \displaystyle \frac{\partial}{\partial x}f,\frac{\partial}{\partial y}f$ exist and are continuous on a neighborhood of $\displaystyle (0,0)$ then $\displaystyle f$ is differentiable there and $\displaystyle f'(0,0)=\text{Jac}_f(0,0)$.