{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe put $h = t v$ in the definition of the derivative.\n\n1. Let $U$, $f$ and $a$ be as in the statement, let $L = Df(a)$, and let $v \\in \\mathbb{R}^n$.\n2. If $v = 0$, then $(f(a + t v) - f(a)) / t = 0 = L v$ for each $t \\ne 0$. So the limit is $L v$. From now on, suppose that $v \\ne 0$.\n3. Since $U$ is [[open|01ZQQDCERP]], there is $\\rho \u003e 0$ such that the [[open ball|0DC3MPR7C9]] $B(a, \\rho)$ lies in $U$. So $a + t v \\in U$ when $|t| \u003c \\rho / \\|v\\|$.\n4. Let $\\varepsilon \u003e 0$. Since $f$ is [[differentiable|0135XZZQ8M]] at $a$, there is $\\delta \u003e 0$ with $\\|f(a + h) - f(a) - L h\\| \u003c \\varepsilon \\|h\\|$. This holds for $0 \u003c \\|h\\| \u003c \\delta$ with $a + h \\in U$.\n5. Let $0 \u003c |t| \u003c \\min(\\delta, \\rho) / \\|v\\|$ and put $h = t v$. Then $a + h \\in U$ by step 3, and $0 \u003c \\|h\\| = |t| \\|v\\| \u003c \\delta$.\n6. Since $L$ is linear, $L h = t L v$. So by step 4,\n$$\\left\\| \\frac{f(a + t v) - f(a)}{t} - L v \\right\\| = \\frac{\\|f(a + h) - f(a) - L h\\|}{|t|} \u003c \\frac{\\varepsilon |t| \\|v\\|}{|t|} = \\varepsilon \\|v\\|.$$\n7. As $\\varepsilon \u003e 0$ was arbitrary, the difference quotient tends to $L v$ as $t \\to 0$.\n8. So $f$ [[has the directional derivative|05EGFBZTGC]] $Df(a)(v)$ at $a$ in the direction $v$.\n\n## Idea\n\nMoving along the line $a + t v$ is a special way of letting $h$ tend to $0$. The derivative controls the error for all small $h$, so in particular for $h = t v$. Dividing by $t$ turns the linear term $L(t v)$ into the constant $L v$.\n",
  "commit": "68a8c3dc73fd55004b2796431c607c30960d1a85",
  "front": {
    "kind": "proof",
    "msc": [
      "26B05"
    ],
    "name": "Proof that the derivative gives the directional derivatives",
    "proves": "0A6PMR20N4",
    "sources": [
      {
        "locator": "Proposition 8.3.9",
        "url": "https://www.jirka.org/ra/realanal2.pdf"
      }
    ],
    "tag": "NY1PHJHKN9",
    "uses": [
      "0135XZZQ8M",
      "05EGFBZTGC",
      "01ZQQDCERP",
      "0DC3MPR7C9"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/NY1PHJHKN9/",
  "links": [
    {
      "field": "proves",
      "target": "0A6PMR20N4"
    },
    {
      "field": "uses",
      "target": "0135XZZQ8M"
    },
    {
      "field": "uses",
      "target": "05EGFBZTGC"
    },
    {
      "field": "uses",
      "target": "01ZQQDCERP"
    },
    {
      "field": "uses",
      "target": "0DC3MPR7C9"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/NY1PHJHKN9.md",
  "msc": [
    "26B05"
  ],
  "name": "Proof that the derivative gives the directional derivatives",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/NY1/NY1PHJHKN9.md",
  "status": null,
  "tag": "NY1PHJHKN9",
  "uri": "m2://NY1PHJHKN9",
  "url": "https://math2.tamnd.com/wiki/proof-that-the-derivative-gives-the-directional-derivatives/",
  "writing": {
    "tag": "NY1PHJHKN9",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}