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I have to prove that the following function is differentiable and to find its derivatives at any point.

$$f: \mathbb{R}^2 \rightarrow \mathbb{R}, (x,y) \mapsto x^2+y^2$$

In my book there is a theorem and one part of it is the following:

Rule of summation:

Let $f:U \subset \mathbb{R}^n \rightarrow \mathbb{R}^m$ and $g:U \subset \mathbb{R}^n \rightarrow \mathbb{R}^m$ be differentiable at $x_0$. Then $h(x)=f(x)+g(x)$ is differentiable at $x_0$ and $Dh(x_0)=Df(x_0)+Dg(x_0)$.

($Df(x_0)$ is the matrix of the partial derivatives of $f$ at $x_0$)

Could we use this to show that the function $f: \mathbb{R}^2 \rightarrow \mathbb{R}, (x,y) \mapsto x^2+y^2$ is differentiable?

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  • $\begingroup$ I'm not sure that's necessarily what they want you to do. You can write down the matrix (vector, in this case) of partial derivatives, and then all you need to do is check that if you subtract off the linear approximation then you get something which decays faster than linearly. Here the linear approximation at $(x_0,y_0)$ is $g(x,y)=x_0^2+y_0^2+2x_0(x-x_0)+2y_0(y-y_0)$, so now subtract that from $x^2+y^2$ and try to bound it. $\endgroup$ – Ian Apr 23 '15 at 23:05
  • $\begingroup$ Seems reasonable, letting $h(x, y) = f(x)+g(y)$. $\endgroup$ – Brian Tung Apr 23 '15 at 23:05
  • $\begingroup$ Hint (in the case you have already seen the composition of differentiable functions is also differentiable) Let $g:\mathbb R\longrightarrow \mathbb R$ be given by $g(x):=x^2$. Notice $f(x, y):=g\circ p_1(x, y)+ g\circ p_2(x, y)$ where $p_j:\mathbb R^2\longrightarrow \mathbb R$ is given by $p_j(x_1, x_2):=x_j$. Hence, it suffices showing $g$ and $p_j$ are differentiable (showing $g$ is differentiable in a point is the same as showing it has a derivative in this point)... $\endgroup$ – PtF Apr 23 '15 at 23:06
  • $\begingroup$ How could we show that $g$ and $p_j$ are differentiable ?? @PtF $\endgroup$ – Mary Star Apr 23 '15 at 23:09
  • $\begingroup$ Well, for $g$ it is easy because $g^\prime$ exists in every $x\in \mathbb R$ hence it is differentiable everywhere. For $p_j$ you might simply compute the partial derivatives and show they are continuous what is easy to be done (again, this is only helpful if you have already seen that if a function have continuous partial derivatives in a point then it is differentiable in that point)..However, I'm trying to avoid to do a lot computations, however it's more likely you must do using @Ian 's idea. $\endgroup$ – PtF Apr 23 '15 at 23:13
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you have $f(x,y) = x^2 + y^2.$ if you look at $$\begin{align}f(x+h, y+k) &= (x+h)^2 + (y+k)^2 \\ &= x^2 + y^2 + 2xh + 2ky + h^2 + k^2 \\ &= f(x,y) + 2xh + 2yk + h^2 + k^2 \end{align}\tag 1$$

taking $$h = \epsilon \cos t, k = \epsilon \sin t$$ in $(1)$ we have, after some rearrangement, $$\frac{f(x + \epsilon \cos t, y+\epsilon \sin t )-f(x,y)}{\epsilon} = 2x\cos t + 2y\sin t+\epsilon \rightarrow 2x\cos t + 2y\sin t \tag 2$$ as $\epsilon \rightarrow 0.$ therefore $$ \text{ the derivative of } x^2 + y^2 \text{ is } (2x, 2y)^\top.$$

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