show that $\partial_u(f \circ g)=f'(g(x))\partial_ug(x)= f'(g(x))(\nabla g(x) \cdot u)$

Question

Let $f :\mathbb{R} \to \mathbb{R}$ be a real-valued differentiable function. If $g :\mathbb{R^3} \to \mathbb{R}$ is differentiable, show that $$\partial_u(f \circ g)=f'(g(x))\partial_ug(x)= f'(g(x))(\nabla g(x) \cdot u)$$ where, $u$ is an unit vector.

I cannot seem to get this to work. It pretty much uses only the chain rule(?), but the notation confuses me a lot. How should I go about this?

Answer 1

It seems that your source uses the notation $$\partial_u\phi(x):=d\phi(x).u\ ,$$ where $u$ is any tangent vector, maybe a unit tangent vector, at $x$. This means that $\partial_u$ is considered as directional derivative in direction $u$.

We now are given a composition $f\circ g$ with $g: \>{\mathbb R}^3\to{\mathbb R}$ and $f:\>{\mathbb R}\to{\mathbb R}$. For such functions one has $$dg(x).u=\nabla g(x)\cdot u,\qquad df(y).v=f'(y) v\ .$$ The chain rule therefore gives $$\eqalign{\partial_u(f\circ g)(x)&=d(f\circ g)(x).u=df\bigl(g(x)\bigr).(dg(x).u\bigr)\cr &=f'\bigl(g(x)\bigr)\bigl(\nabla g(x)\cdot u\bigr)\ .\cr}$$

Answer 2

A matricial view helps to settle your confusion: for a composition $$\mathbb R^3\stackrel{g}\to\mathbb R\stackrel{f}\to\mathbb R$$ then generically $(f\circ g)'=f'(g)\cdot g'$ is the chain's rule, and particularization would be $$\nabla (f\circ g)=f'(g)\cdot\nabla g.$$ If use $x,y,z$ as the coordinated function on $\mathbb R^3$ and $t$ on $\mathbb R$ then $$ \left[\begin{array}{ccc} \frac{\partial(f\circ g)}{\partial x} & \frac{\partial(f\circ g)}{\partial y} & \frac{\partial(f\circ g)}{\partial z} \end{array}\right]= \frac{df}{dt}(g) \left[\begin{array}{ccc} \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} & \frac{\partial g}{\partial z} \end{array} \right]$$ which implies $$\frac{\partial(f\circ g)}{\partial x}=\frac{df}{dt}(g) \frac{\partial g}{\partial x} \quad,\quad \frac{\partial(f\circ g)}{\partial y}=\frac{df}{dt}(g) \frac{\partial g}{\partial y} \quad,\quad \frac{\partial(f\circ g)}{\partial z}=\frac{df}{dt}(g) \frac{\partial g}{\partial z} .$$ But if you want a directional derivative in the direction $\mathbf u=u^1\vec i+u^2\vec j+u^3\vec k$ then $$\frac{\partial(f\circ g)}{\partial\mathbf u}= u^1\frac{df}{dt}(g) \frac{\partial g}{\partial x}+ u^2\frac{df}{dt}(g) \frac{\partial g}{\partial y}+ u^3\frac{df}{dt}(g) \frac{\partial g}{\partial z}.$$