1

I want to prove

Suppose $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is differentiable, and let $v$ be a unit vector in $\mathbb{R}^n$. Prove that the directional derivative of $f$ at point $\bar{x}$ with respect to direction $v$ satisfies$$f'(\bar{x};v)=\nabla f(\bar{x})\cdot v$$ Here $\nabla f$ is the gradient of $f$.

I've already done the following: Suppose $\bar{x}=(\bar{x}_1,\cdots,\bar{x}_n)$, and $v=(v_1,\cdots,v_n)$, and $$\mathrm{RHS}=\dfrac{\partial f}{\partial x_1}(\bar{x}_1)v_1+\cdots+\dfrac{\partial f}{\partial x_n}(\bar{x}_n)v_n.$$ $$\mathrm{LHS}=\lim_{t\downarrow 0}\dfrac{f(\bar{x}+tv)-f(\bar{x})}{t}\\=\lim_{t\downarrow0}\dfrac{f(\bar{x}_1+tv_1,\cdots,\bar{x}_n+tv_n)-f(\bar{x}_1,\cdots,\bar{x}_n)}{t}$$ However I don't know how to do the next steps to prove the fact. Can anyone do me a favour in fixing the proof?

Scanners
  • 188

2 Answers2

1

HINT

Define $F(t):=f(\bar{x}_1+tv_1,\cdots,\bar{x}_n+t v_n)$. Can you express $F'(t)$ in two different ways?

Tito Eliatron
  • 6,612
1

For $t \in \mathbb R$ define $g(t):=f(\bar{x}+tv).$

Then $g$ is differentiable and , by the chain rule:

$$g'(t)=\nabla f(\bar{x}+tv)\cdot v.$$

On the other hand:

$$f'(\bar{x};v)=\lim_{t \to 0}\dfrac{f(\bar{x}+tv)-f(\bar{x})}{t}=g'(0).$$

Fred
  • 77,394