Prove that $f$ and $g$ are differentiable on $\Bbb{R}^n$ and $\Bbb{R}^n-\{0\}$, respectively and compute $f'(x)$ and $g'(x)$

Question

Let $f,g:\Bbb{R}^n\to\Bbb{R} $ be defined by $f(x)=\langle x,x\rangle$ and $g(x)=\sqrt{\langle x,x\rangle}\,,$ respectively.

Prove that $f$ and $g$ are differentiable on $\Bbb{R}^n$ and $\Bbb{R}^n-\{0\}$, respectively and compute $f'(x)$ and $g'(x).$

If I may ask, does this $\langle x,x\rangle$ also mean inner product? If yes, my problem is half solved. If no, I would need some nice proofs.

Answer 1

$f$ is differentiable in the Frechet sense in $a$ if there exists a function $\phi: \mathbb{R^n}\mapsto \mathcal{L}( \mathbb{R^n}, \mathbb{R}) $ such that: $$f( a + h) = f(a) + \phi(a).h + o(h)$$

When $h$ approaches $0$.

In our case, $f(a+h) = \langle a+h, a+h\rangle = \langle a, a\rangle + 2\langle a, h \rangle + \langle h, h\rangle$.
Since $\langle h, h\rangle = o(h)$, $f$ is differentiable over $\mathbb{R}^n$ and we have $f'(a).h = \langle 2a, h\rangle$.

This means $f'(a) = \langle 2a, \cdot \rangle$.
Using the jacobian matrix, we have $J_f(a) = 2a^T$.
In terms of gradient, we have $\nabla f(a) = 2a$.

Now, to get $g'$ you can apply the chain rule to $g = \sqrt\cdot \circ f$.
Since $\sqrt \cdot$ is derivable over $\mathbb{R^*}$, and $f^{-1}(\{0_{\mathbb{R}}\}) = \{0_{\mathbb{R}^n}\}$, $g$ is differentiable over $\mathbb{R^n}-\{0\}$ and:

$$g'(a) = \frac{1}{2\sqrt {f(a)}} f'(a)$$

Answer 2

Let $x = (x_1, \ldots, x_n) \in \mathbb{R}^n$. Then $$ f(x) = \langle x, x \rangle = x_1^2 + \ldots + x_n^2. $$ Since $f$ is just a multivariable polynomial, it is differentiable. The partial derivatives are $$ \frac{\partial f}{\partial x_j} = 2x_j, $$ and the derivative is the $1 \times n$ matrix (also known as the gradient) $$ f'(x) = \begin{pmatrix}\frac{\partial f}{\partial x_1} & \ldots & \frac{\partial f}{\partial x_n}\end{pmatrix} = \begin{pmatrix}2x_1 & \ldots & 2x_n\end{pmatrix}. $$

For $g$, note that $ \langle x,x \rangle \geq 0 $ with equality if and only if $x = 0$, and since $y \mapsto \sqrt{y}$ is differentiable for $y > 0$, the chain rule gives us that $x \mapsto \sqrt{\langle x,x\rangle}$ is differentiable on $\mathbb{R}^n-\{0\}$.

Can you compute the derivative of $g$ on your own now?