Is $f(x)=|x|^2$ a $C^1$ function?

Question

$\bigtriangledown f(x) = 2|x|$ is continuous, so based on $C^1$ function definition $f(x)$ is $C^1$ function.

Can someone help me to confirm this? Thanks in advance.

Answer 1

Notice that $$f(x)=|x|^2=x^2$$

It is a polynomial function, hence it is a $C^1$ function.

Remark:

The compute $\nabla f$ that you computed is nonnegative, which is not the case. The derivative should be $2x$.

Answer 2

Note that $f(x)=\displaystyle\sum_{i=1}^{n}x_{i}^{2}$, so $\dfrac{\partial f}{\partial x_{i}}(x)=2x_{i}$ is continuous for each $i=1,2,...,n$.