Vtyjkyuk

Let $f,g:BbbR^ntoBbbR $ be defined by $f(x)=langle x,xrangle$ and $g(x)=sqrtlangle x,xrangle,,$ respectively.

Prove that $f$ and $g$ are differentiable on $BbbR^n$ and $BbbR^n-0$, respectively and compute $f'(x)$ and $g'(x).$

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

$f$ is differentiable in the Frechet sense in $a$ if there exists a function $phi: mathbbR^nmapsto mathcalL( mathbbR^n, mathbbR) $ such that: $$f(
a + h) = f(a) + phi(a).h + o(h)$$

When $h$ approaches $0$.

In our case, $f(a+h) = langle a+h, a+hrangle = langle a, arangle + 2langle a, h rangle + langle h, hrangle$.

Since $langle h, hrangle = o(h)$, $f$ is differentiable over $mathbbR^n$ and we have $f'(a).h = langle 2a, hrangle$.

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 $g = sqrtcdot circ f$.

Since $sqrt cdot$ is derivable over $mathbbR^*$, and $f^-1(0_mathbbR) = 0_mathbbR^n$, $g$ is differentiable over $mathbbR^n-0$ and:

$$g'(a) = frac12sqrt f(a) f'(a)$$

Let $x = (x_1, ldots, x_n) in mathbbR^n$. 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
$$ fracpartial fpartial x_j = 2x_j, $$
and the derivative is the $1 times n$ matrix (also known as the gradient)
$$ f'(x) = beginpmatrixfracpartial fpartial x_1 & ldots & fracpartial fpartial x_nendpmatrix = beginpmatrix2x_1 & ldots & 2x_nendpmatrix. $$

For $g$, note that $ langle x,x rangle geq 0 $
with equality if and only if $x = 0$, and since $y mapsto sqrty$ is differentiable for $y > 0$, the chain rule gives us that $x mapsto sqrtlangle x,xrangle$ is differentiable on $mathbbR^n-0$.

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