Directional derivative of $f=f(\nabla \cdot\mathbf{u})$

Question

How do you evaluate the directional derivative of

$$f=f(\nabla \cdot\mathbf{u})\tag{1}$$

I've tried this but I'm not sure that my answer is correct, here is my attempt:

The definition of the directional derivative:

$$Df(\mathbf{u})[\mathbf{u^*}]=\lim_{\epsilon\rightarrow 0}\left[\frac{df(\mathbf{u}+\epsilon\mathbf{u^*})}{d\epsilon}\right]\tag{2}$$

Attempt 1

$$Df(\nabla\cdot \mathbf{u})[\mathbf{u}^*]=\lim_{\epsilon\rightarrow 0}\frac{f(\nabla\cdot(\mathbf{u}+\mathbf{u}^*\epsilon))}{d\epsilon}$$

By the chain rule: $$=\lim_{\epsilon\rightarrow 0}\left[\frac{df(\nabla\cdot(\mathbf{u}+\mathbf{u}^*\epsilon))}{d(\nabla\cdot(\mathbf{u}+\mathbf{u}^*\epsilon))}\frac{d\nabla\cdot(\mathbf{u}+\mathbf{u}^*\epsilon)}{d\epsilon}\right]$$

$$=\lim_{\epsilon\rightarrow 0}\left[\frac{df(\nabla\cdot(\mathbf{u}+\mathbf{u}^*\epsilon))}{d(\nabla\cdot(\mathbf{u}+\mathbf{u}^*\epsilon))}\nabla\cdot\mathbf{u}^*\right]$$

$$=\lim_{\epsilon\rightarrow 0}\left[f'(\nabla \cdot(\mathbf{u}+\mathbf{u}^*\epsilon))\nabla\cdot\mathbf{u}^*\right]$$

$$=f'(\nabla \cdot\mathbf{u})\nabla\cdot\mathbf{u}^*\tag{3}$$

Attempt 2

I really don't think this is right but the notation of the question I've been given (which is $(1)$) is confusing me:

$$Df(\nabla\cdot \mathbf{u})[\mathbf{u}^*]=\lim_{\epsilon\rightarrow 0}\frac{f(\nabla\cdot(\mathbf{u})+\mathbf{u}^*\epsilon)}{d\epsilon}\tag{4}$$

$$=f'(\nabla \cdot\mathbf{u})\mathbf{u}^*\tag{5}$$

Are either of these attempts correct? And if not how do you derive the correct answer?

Answer 1

If $u: \mathbb{R}^3 \to \mathbb{R}^3$ is a fixed vector-valued function, then define $g:\mathbb{R}^3 \to \mathbb{R}$ by $g(x,y,z)=\nabla \cdot u(x,y,z)$. Then $F(x,y,z)=f(\nabla \cdot u) = (f\circ g) (x,y,z)$, where $f:\mathbb{R}\to \mathbb{R}$. The directional derivative of $F$ in the $v$-direction is denoted by $D_v F$ and is a real-valued function $\mathbb{R}^3 \to \mathbb{R}$. Then $$D_v F(x,y,z) = D(F(x,y,z))(v) = D(f\circ g(x,y,z))(v)= [Df(g(x,y,z))\circ Dg(x,y,z)](v) = f'(g(x,y,z))[\nabla g(x,y,z) \cdot v] $$ Thus $D_v F(x,y,z) = f'(\nabla \cdot u)[\nabla(\nabla \cdot u) \cdot v]$