differentiability multivariable function

**storchfire1X** · May 9th 2011, 05:33 PM

Hi, I need help on the following:

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

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

Thanks

**TheEmptySet** · May 9th 2011, 07:02 PM

Originally Posted by storchfire1X

Hi, I need help on the following:

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

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

Thanks

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

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

If you evaluate this at zero we get

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

Now you need to calculate the limit

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

and show that it is equal to zero!

Where

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

This should get you started.

**Drexel28** · May 9th 2011, 09:23 PM

Originally Posted by storchfire1X

Hi, I need help on the following:

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

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

Thanks

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

Thread: differentiability multivariable function

LinkBack

Thread Tools

Display

differentiability multivariable function

Similar Math Help Forum Discussions

Differentiability of the Weierstrauss Function

Differentiability, multivariable calculus

Differentiability and the chain rule (multivariable calculus)

Even/odd function differentiability

Differentiability of the function

Search Tags