# Show that f is differentiable

• Dec 15th 2009, 08:19 AM
thaopanda
Show that f is differentiable
Let f : $\displaystyle R^3 \rightarrow R^2$ be given by f ($\displaystyle x_1, x_2, x_3$) = ($\displaystyle x_3, x_1$). Show that f is differentiable at $\displaystyle X_0$ = (0,0,0) and find df($\displaystyle X_0$).

I figured to use the definition so:

f(X) = f($\displaystyle X_0$) + df($\displaystyle X_0$)(X-$\displaystyle X_0$) + o(|x-$\displaystyle X_0$|) as X $\displaystyle \rightarrow X_0$

f($\displaystyle X_0$) goes to 0 and X-$\displaystyle X_0$ is just X
So I'm left with:
f(X) = df($\displaystyle X_0$)(X) + o(|X|) as X $\displaystyle \rightarrow X_0$

But I'm not sure how to find df($\displaystyle X_0$)
Doing so would that f is differentiable at $\displaystyle X_0$ and find it at the same time, yes?
• Dec 15th 2009, 09:14 AM
Opalg
Quote:

Originally Posted by thaopanda
Let f : $\displaystyle R^3 \rightarrow R^2$ be given by f ($\displaystyle x_1, x_2, x_3$) = ($\displaystyle x_3, x_1$). Show that f is differentiable at $\displaystyle X_0$ = (0,0,0) and find df($\displaystyle X_0$).

I figured to use the definition so:

f(X) = f($\displaystyle X_0$) + df($\displaystyle X_0$)(X-$\displaystyle X_0$) + o(|x-$\displaystyle X_0$|) as X $\displaystyle \rightarrow X_0$

f($\displaystyle X_0$) goes to 0 and X-$\displaystyle X_0$ is just X
So I'm left with:
f(X) = df($\displaystyle X_0$)(X) + o(|X|) as X $\displaystyle \rightarrow X_0$

But I'm not sure how to find df($\displaystyle X_0$)
Doing so would that f is differentiable at $\displaystyle X_0$ and find it at the same time, yes?

The correct way to think of differentiation is that it is a linearisation process. In elementary calculus, you have a function f(x), whose graph is some sort of curve, and when you differentiate it you approximate the curve by a straight line, namely the tangent to the curve.

For a function $\displaystyle f:\mathbb{R}^3\to\mathbb{R}^2$, when you differentiate it you are approximating f by a linear map from $\displaystyle \mathbb{R}^3$ to $\displaystyle \mathbb{R}^2$. But linear maps between vector spaces are described by matrices. For a linear map from $\displaystyle \mathbb{R}^3$ to $\displaystyle \mathbb{R}^2$, you write the elements of $\displaystyle \mathbb{R}^3$ and $\displaystyle \mathbb{R}^2$ as column vectors, and the linear map is given by a $\displaystyle 2\times3$ matrix. The formula for the derivative matrix $\displaystyle Df(X_0)$ is that it satisfies the equation $\displaystyle f(X) = f(X_0) + Df(X_0)(X-X_0) + o(|X-X_0|)$.

For this example, f takes the vector $\displaystyle X = \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\in\mathb b{R}^3$ to the vector $\displaystyle f(X) = \begin{bmatrix}x_3\\x_1\end{bmatrix}\in\mathbb{R}^ 2$. You want to find Df(0), which will be a $\displaystyle 2\times3$ matrix such that $\displaystyle \begin{bmatrix}x_3\\x_1\end{bmatrix} = Df(0)\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} + o(|X|)$. As it happens, this function f is already linear, so the $\displaystyle o(|X|)$ term disappears, and $\displaystyle \begin{bmatrix}x_3\\x_1\end{bmatrix} = \begin{bmatrix}0&0&1\\1&0&0\end{bmatrix}\begin{bma trix}x_1\\x_2\\x_3\end{bmatrix}$. So $\displaystyle Df(0) = \begin{bmatrix}0&0&1\\1&0&0\end{bmatrix}$.

As you already said, this process of finding $\displaystyle Df(X_0)$ automatically includes a proof that f is differentiable at $\displaystyle X_0$.
• Dec 15th 2009, 09:51 AM
thaopanda
Thank you very much! That made it very clear for me finally (Happy)