The Lagrangian is defined as follows:
$$\mathcal{L}(x,\mu)=\lVert x \rVert^2_2+\underbrace{\mu^T(y-Hx)}_{**}$$ Then the derivatives of the lagrangian is taken which is what I don't understand:
$\dfrac\partial{\partial x} \mathcal{L}(x)=2x-H^T\mu$
$\dfrac\partial{\partial \mu}\mathcal{L}(x)=y-Hx$
How is the derivative of the second term ($**$) calculated both w.r.t $\mu$ and $x$?
Go from first principles and think of derivatives as linear mappings (see Cartan's Calculus in Banach spaces). Let $\langle , \rangle$ be the inner product that induces the norm $|| \cdot ||$ you are using. The term in question is then: $$ G(x, \mu) = \langle \mu, y - Hx \rangle $$ As a step toward computing derivatives, let's first examine the finite differences $G(x + \triangle x, \mu) - G(x, \mu)$ and $G(x, \mu + \triangle \mu)- G(x, \mu)$. By direct computation, $$ G(x + \triangle x, \mu) = \langle \mu, y - Hx \rangle - \langle \mu, H \triangle x \rangle, $$ so $$ G(x + \triangle x, \mu) - G(x, \mu) = - \langle \mu, H \triangle x \rangle. $$ In other words, the derivative of $G$ with respect to $x$ at $x$ is the linear mapping that sends $\triangle x$ to $- \langle \mu, H \triangle x \rangle$. The way to write this mapping without using the scalar product is by using transposition: $- H^{T} \mu$.
Similarly for differentiating w.r.t. $\mu$.