{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe apply the rule for the derivative of an inverse function to the tangent on $(-\\pi/2, \\pi/2)$.\n\n1. Let $I = (-\\pi/2, \\pi/2)$, and let $T \\colon I \\to \\R$ be the restriction of the [[tangent function|06T0HS14PP]] to $I$.\n2. No point of $I$ has the form $\\frac{\\pi}{2} + k\\pi$ with $k \\in \\Z$. So $\\cos y \\ne 0$ for all $y \\in I$, and $T$ is defined on all of $I$.\n3. By the [[derivative of tangent|0SAMR34WXC]], $T$ is differentiable at each $y \\in I$, with $T'(y) = 1 + \\tan^2 y$.\n4. By step 3, $T$ is continuous on $I$, since [[differentiable functions are continuous|07ZMX4ANJS]].\n5. For each $y \\in I$, step 3 gives $T'(y) = 1 + \\tan^2 y \\ge 1 \u003e 0$.\n6. By steps 4 and 5, $T$ is strictly increasing on $I$, since [[a function with positive derivative is strictly increasing|07J5X2T1ZX]].\n7. By the definition of the [[arctangent|065XW2V6C3]], each real $x$ has exactly one $y \\in I$ with $T(y) = x$, namely $y = \\arctan x$.\n8. So $T \\colon I \\to \\R$ is a bijection, and its [[inverse function|0W766JAPSN]] is $T^{-1} = \\arctan$.\n9. Fix a real number $x$, and let $a = \\arctan x \\in I$. Then $T(a) = x$, and $T'(a) = 1 + \\tan^2 a = 1 + x^2 \\ne 0$.\n10. The interval $I$ is open, and $T$ is continuous and strictly monotone on it by steps 4 and 6. So the [[derivative of an inverse function|0M0RZ3HSB0]] applies at $a$.\n11. It gives that $\\arctan = T^{-1}$ is differentiable at $x = T(a)$, with\n$$\\arctan'(x) = \\frac{1}{T'(a)} = \\frac{1}{1 + x^2}.$$\n12. Since $x$ was an arbitrary real number, $\\arctan$ is differentiable on $\\R$ with $\\frac{d}{dx} \\arctan x = \\frac{1}{1 + x^2}$.\n\n## Idea\n\nThe arctangent undoes the tangent on $(-\\pi/2, \\pi/2)$, so its slope at $x$ is the reciprocal of the slope of $\\tan$ at $\\arctan x$. That slope is $1 + \\tan^2 y$, and $\\tan y = x$ there. So the answer depends on $x$ alone: $\\frac{1}{1 + x^2}$.\n",
  "commit": "68a8c3dc73fd55004b2796431c607c30960d1a85",
  "front": {
    "kind": "proof",
    "msc": [
      "33B10",
      "26A24"
    ],
    "name": "Proof of the derivative of arctangent",
    "proves": "05P4B4Q22V",
    "sources": [
      {
        "locator": "Differentiating the inverse tangent function",
        "url": "https://en.wikipedia.org/wiki/Differentiation_of_trigonometric_functions"
      }
    ],
    "tag": "DT9HKT4XTE",
    "uses": [
      "065XW2V6C3",
      "06T0HS14PP",
      "0SAMR34WXC",
      "07ZMX4ANJS",
      "07J5X2T1ZX",
      "0W766JAPSN",
      "0M0RZ3HSB0"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/DT9HKT4XTE/",
  "links": [
    {
      "field": "proves",
      "target": "05P4B4Q22V"
    },
    {
      "field": "uses",
      "target": "065XW2V6C3"
    },
    {
      "field": "uses",
      "target": "06T0HS14PP"
    },
    {
      "field": "uses",
      "target": "0SAMR34WXC"
    },
    {
      "field": "uses",
      "target": "07ZMX4ANJS"
    },
    {
      "field": "uses",
      "target": "07J5X2T1ZX"
    },
    {
      "field": "uses",
      "target": "0W766JAPSN"
    },
    {
      "field": "uses",
      "target": "0M0RZ3HSB0"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/DT9HKT4XTE.md",
  "msc": [
    "33B10",
    "26A24"
  ],
  "name": "Proof of the derivative of arctangent",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/DT9/DT9HKT4XTE.md",
  "status": null,
  "tag": "DT9HKT4XTE",
  "uri": "m2://DT9HKT4XTE",
  "url": "https://math2.tamnd.com/wiki/proof-of-the-derivative-of-arctangent/",
  "writing": {
    "tag": "DT9HKT4XTE",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}