{
  "api_version": "v1",
  "publisher": {
    "name": "Manazoo",
    "url": "https://manazoo.com/"
  },
  "license": "All rights reserved（AI の参照・要約と出典リンクつきの引用は可。転載・再配布・販売は不可）",
  "terms_url": "https://manazoo.com/terms/",
  "disclaimer": "単元の並びは学習指導要領に沿って作り、年号や人物などの事実は公開データと照らし合わせています。誤りに気づいたら意見箱からお知らせください。確かめて直します。テストや受験の前には、教科書や先生の説明もあわせて確認してください。",
  "attribution": "事実データの一部は Wikidata（CC0）による。単元の構成は文部科学省「小学校学習指導要領」「中学校学習指導要領」「高等学校学習指導要領」（政府標準利用規約2.0）。",
  "kind": "quiz",
  "unit": {
    "slug": "differential-calculus",
    "title": "微分法",
    "url": "https://manazoo.com/high/math-3/differential-calculus/"
  },
  "canonical_url": "https://manazoo.com/high/math-3/differential-calculus/quiz/",
  "questions": [
    {
      "id": "q-cfd-chain-rule-value-at-1",
      "type": "input",
      "prompt": "y＝(3x＋2)⁴の導関数dy/dxを合成関数の公式で求め、x＝0でのdy/dxの値を求めよ。",
      "choices": null,
      "answer": "96",
      "accept": [],
      "explanation": "u＝3x＋2とおくとdy/dx＝4u³×3＝12(3x＋2)³。x＝0ではu＝2なので、dy/dx＝12×2³＝96。",
      "entities": [
        "composite-function-derivative"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-cfd-inner-outer-choice",
      "type": "choice",
      "prompt": "y＝(x²＋1)³で、u＝x²＋1とおいたとき、x＝1でのdy/dxの値はどれか。",
      "choices": [
        "6",
        "12",
        "24",
        "48"
      ],
      "answer": 2,
      "accept": [],
      "explanation": "dy/dx＝3u²×2x＝6x(x²＋1)²。x＝1ではu＝2なので、dy/dx＝6×1×2²＝24。",
      "entities": [
        "composite-function-derivative"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-cfd-inverse-sqrt-value-at-4",
      "type": "input",
      "prompt": "y＝√xの導関数dy/dxをx＝y²の逆関数として求め、x＝4でのdy/dxの値を求めよ。分数の形で、/（スラッシュ）を使って答えよう。",
      "choices": null,
      "answer": "1/4",
      "accept": [
        "1/4"
      ],
      "explanation": "dy/dx＝1/(2√x)なので、x＝4では1/(2×2)＝1/4。",
      "entities": [
        "composite-function-derivative"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-cfd-rational-power-choice",
      "type": "choice",
      "prompt": "(xʳ)′＝rxʳ⁻¹がrが有理数のときも成り立つとすると、f(x)＝\\(x^{1/3}\\)の導関数f′(x)はどれか。",
      "choices": [
        "\\((1/3)x^{-2/3}\\)",
        "\\((1/3)x^{2/3}\\)",
        "3x²",
        "\\(x^{-2/3}\\)"
      ],
      "answer": 0,
      "accept": [],
      "explanation": "r＝1/3を(xʳ)′＝rxʳ⁻¹に当てはめると、f′(x)＝\\((1/3)x^{1/3-1} = (1/3)x^{-2/3}\\)になる。",
      "entities": [
        "composite-function-derivative"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-cip-concavity-sign-choice",
      "type": "choice",
      "prompt": "f″(x)＞0の区間で、グラフの凹凸はどうなるか。",
      "choices": [
        "増加",
        "下に凸",
        "上に凸",
        "減少"
      ],
      "answer": 1,
      "accept": [],
      "explanation": "f″(x)＞0は接線の傾きが増えていくことを表し、このときグラフは下に凸（谷のような形）になる。",
      "entities": [
        "concavity-and-inflection-points"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-cip-inflection-x-value",
      "type": "input",
      "prompt": "f(x)＝x³−3x²の変曲点のx座標を求めよ。",
      "choices": null,
      "answer": "1",
      "accept": [],
      "explanation": "f″(x)＝6x−6＝0を解くとx＝1。x＝1の前後でf″の符号が変わるので、x＝1が変曲点のx座標になる。",
      "entities": [
        "concavity-and-inflection-points"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-cip-inflection-y-value-choice",
      "type": "choice",
      "prompt": "f(x)＝x³−3x²の変曲点はどれか。",
      "choices": [
        "(1, −2)",
        "(2, −4)",
        "(0, 0)",
        "(1, 0)"
      ],
      "answer": 0,
      "accept": [],
      "explanation": "f″(x)＝0となるx＝1で凹凸が切り替わる。f(1)＝1−3＝−2なので、変曲点は(1, −2)。",
      "entities": [
        "concavity-and-inflection-points"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-cip-second-derivative-at-2",
      "type": "input",
      "prompt": "f(x)＝x³−3x²の第二次導関数f″(x)を求め、f″(2)の値を求めよ。",
      "choices": null,
      "answer": "6",
      "accept": [],
      "explanation": "f′(x)＝3x²−6xなので、f″(x)＝6x−6。f″(2)＝6×2−6＝6。",
      "entities": [
        "concavity-and-inflection-points"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-dpq-differentiable-choice",
      "type": "choice",
      "prompt": "f(x)＝|x|がx＝0で微分可能でない理由として正しいものはどれか。",
      "choices": [
        "x＝0の前後で平均変化率の極限が−1と1に分かれ、1つに決まらないから",
        "x＝0でf(x)の値が負になるから",
        "x＝0でf(x)が連続でないから",
        "x＝0でf(x)の値が定義されていないから"
      ],
      "answer": 0,
      "accept": [],
      "explanation": "f(x)＝|x|はx＝0でもつながっていて連続だが、x＜0側の傾きは−1、x＞0側の傾きは1で、hを0に近づける平均変化率の極限が一方向に決まらない。だから微分可能ではない。",
      "entities": [
        "differentiability-and-product-quotient-rule"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-dpq-negative-power-rule-choice",
      "type": "choice",
      "prompt": "商の導関数の公式を使って求めると、f(x)＝x⁻²（＝1/x²）の導関数f′(x)はどれか。",
      "choices": [
        "x⁻³",
        "−2x⁻³",
        "−2x⁻¹",
        "2x⁻³"
      ],
      "answer": 1,
      "accept": [],
      "explanation": "(xⁿ)′＝nxⁿ⁻¹はnが負の整数でも成り立つので、(x⁻²)′＝−2x⁻³になる。f(x)＝1/x²を商の導関数（分子が1で導関数0）で確かめても同じ結果が得られる。",
      "entities": [
        "differentiability-and-product-quotient-rule"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-dpq-product-rule-value-at-1",
      "type": "input",
      "prompt": "f(x)＝(2x＋1)(x²−3)の導関数f′(x)を積の導関数の公式で求め、f′(1)の値を求めよ。",
      "choices": null,
      "answer": "2",
      "accept": [],
      "explanation": "積の導関数の公式を使うと、f′(x)＝6x²＋2x−6になる。f′(1)＝6＋2−6＝2。",
      "entities": [
        "differentiability-and-product-quotient-rule"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-dpq-quotient-rule-value-at-1",
      "type": "input",
      "prompt": "g(x)＝x²/(x＋1)の導関数g′(x)を商の導関数の公式で求め、g′(1)の値を求めよ。分数の形で、/（スラッシュ）を使って答えよう。",
      "choices": null,
      "answer": "3/4",
      "accept": [
        "3/4"
      ],
      "explanation": "g′(x)＝(2x(x＋1)−x²)/(x＋1)²＝(x²＋2x)/(x＋1)²なので、g′(1)＝(1＋2)/4＝3/4。",
      "entities": [
        "differentiability-and-product-quotient-rule"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-eld-e-value-choice",
      "type": "choice",
      "prompt": "nを限りなく大きくしたとき、(1＋1/n)ⁿの値はどれに近づくか。",
      "choices": [
        "約2.71828（e）",
        "1",
        "2",
        "限りなく大きくなる"
      ],
      "answer": 0,
      "accept": [],
      "explanation": "n＝10・100・1000と大きくするほど、(1＋1/n)ⁿは約2.594→2.705→2.717と、eという一定の値に近づいていく。",
      "entities": [
        "exponential-and-logarithmic-derivatives"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-eld-exp-derivative-value-at-0",
      "type": "input",
      "prompt": "y＝\\(e^{2x}\\)の導関数dy/dxを求め、x＝0でのdy/dxの値を求めよ。",
      "choices": null,
      "answer": "2",
      "accept": [],
      "explanation": "dy/dx＝\\(2e^{2x}\\)で、x＝0ではe⁰＝1なので、dy/dx＝2×1＝2。",
      "entities": [
        "exponential-and-logarithmic-derivatives"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-eld-general-exponent-choice",
      "type": "choice",
      "prompt": "f(x)＝\\(2^x\\)の導関数f′(x)を\\((a^x)' = a^x \\log a\\)の公式で求めるとどれか。",
      "choices": [
        "\\(2^x\\)",
        "\\(\\frac{2^x}{\\log 2}\\)",
        "\\(x \\cdot 2^{x-1}\\)",
        "\\(2^x \\log 2\\)"
      ],
      "answer": 3,
      "accept": [],
      "explanation": "\\((a^x)' = a^x \\log a\\)にa＝2を当てはめると、f′(x)＝\\(2^x \\log 2\\)になる。",
      "entities": [
        "exponential-and-logarithmic-derivatives"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-eld-log-derivative-value-at-2",
      "type": "input",
      "prompt": "f(x)＝log(3x)の導関数f′(x)を求め、f′(2)の値を求めよ。分数の形で、/（スラッシュ）を使って答えよう。",
      "choices": null,
      "answer": "1/2",
      "accept": [
        "1/2"
      ],
      "explanation": "log(3x)＝log3＋log xと分けられ、log3は定数なのでf′(x)＝1/x。f′(2)＝1/2。",
      "entities": [
        "exponential-and-logarithmic-derivatives"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-tfd-cosine-derivative-choice",
      "type": "choice",
      "prompt": "cos xの導関数(cos x)′はどれか。",
      "choices": [
        "cos x",
        "−cos x",
        "−sin x",
        "sin x"
      ],
      "answer": 2,
      "accept": [],
      "explanation": "cos x＝sin(x＋π/2)であることと(sin x)′＝cos xから、(cos x)′＝−sin xになる。",
      "entities": [
        "trigonometric-function-derivatives"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-tfd-linear-combo-value-at-0",
      "type": "input",
      "prompt": "y＝3sin x−2cos xの導関数dy/dxを求め、x＝0でのdy/dxの値を求めよ。",
      "choices": null,
      "answer": "3",
      "accept": [],
      "explanation": "dy/dx＝3cos x＋2sin xなので、x＝0ではcos0＝1、sin0＝0より、dy/dx＝3×1＋2×0＝3。",
      "entities": [
        "trigonometric-function-derivatives"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-tfd-product-rule-choice",
      "type": "choice",
      "prompt": "f(x)＝sin x cos xの導関数f′(x)を積の導関数の公式で求めるとどれか。",
      "choices": [
        "cos2x",
        "2sin x cos x",
        "−sin2x",
        "sin2x"
      ],
      "answer": 0,
      "accept": [],
      "explanation": "f′(x)＝(sin x)′cos x＋sin x(cos x)′＝cos²x−sin²x。これは2倍角の公式でcos2xと等しい。",
      "entities": [
        "trigonometric-function-derivatives"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-tfd-tangent-value-at-pi4",
      "type": "input",
      "prompt": "f(x)＝tan xの導関数f′(x)＝1/cos²xを使って、f′(π/4)の値を求めよ。",
      "choices": null,
      "answer": "2",
      "accept": [],
      "explanation": "cos(π/4)＝1/√2なので、cos²(π/4)＝1/2。f′(π/4)＝1/(1/2)＝2。",
      "entities": [
        "trigonometric-function-derivatives"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-tnl-horizontal-tangent-choice",
      "type": "choice",
      "prompt": "曲線上の点(a, f(a))でf′(a)＝0のとき、法線の方程式はどれか。",
      "choices": [
        "y＝0",
        "y＝−x＋a",
        "y＝f(a)",
        "x＝a"
      ],
      "answer": 3,
      "accept": [],
      "explanation": "f′(a)＝0のとき接線はy＝f(a)（横の直線）になり、それに垂直な法線は縦の直線x＝aになる。",
      "entities": [
        "tangent-and-normal-lines"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-tnl-normal-intercept-log-at-1",
      "type": "input",
      "prompt": "曲線y＝log xの点(1, 0)における法線の方程式はy＝−x＋kの形になる。kの値を求めよ。",
      "choices": null,
      "answer": "1",
      "accept": [],
      "explanation": "f′(1)＝1/1＝1なので、法線の傾きは−1。y−0＝−1(x−1)より、y＝−x＋1。k＝1。",
      "entities": [
        "tangent-and-normal-lines"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-tnl-normal-slope-exp-at-0",
      "type": "input",
      "prompt": "曲線y＝\\(e^x\\)の点(0, 1)における法線の傾きを求めよ。",
      "choices": null,
      "answer": "-1",
      "accept": [
        "−1"
      ],
      "explanation": "f′(0)＝e⁰＝1なので、法線の傾きは−1/f′(0)＝−1/1＝−1。",
      "entities": [
        "tangent-and-normal-lines"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-tnl-tangent-sine-choice",
      "type": "choice",
      "prompt": "曲線y＝sin xの点(0, 0)における接線の方程式はどれか。",
      "choices": [
        "y＝x",
        "y＝−x",
        "y＝1",
        "x＝0"
      ],
      "answer": 0,
      "accept": [],
      "explanation": "f′(0)＝cos0＝1なので、接線はy−0＝1(x−0)よりy＝x。",
      "entities": [
        "tangent-and-normal-lines"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-vaa-acceleration-at-2",
      "type": "input",
      "prompt": "位置x(t)＝t³−6t²＋9tで動く点の加速度a(t)を求め、a(2)の値を求めよ。",
      "choices": null,
      "answer": "0",
      "accept": [],
      "explanation": "v(t)＝3t²−12t＋9なので、a(t)＝v′(t)＝6t−12。a(2)＝6×2−12＝0。",
      "entities": [
        "velocity-acceleration-and-approximation"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-vaa-approximation-meaning-choice",
      "type": "choice",
      "prompt": "近似式Δy≈f′(x)Δxが成り立つ理由として正しいものはどれか。",
      "choices": [
        "yの値がxの値に比例するから",
        "xの近くでは曲線が接線にほとんど重なって見えるから",
        "f′(x)が必ず一定の値になるから",
        "Δxがどんな大きさでも誤差が0になるから"
      ],
      "answer": 1,
      "accept": [],
      "explanation": "接線の方程式y−f(a)＝f′(a)(x−a)で、x−a＝Δx、y−f(a)＝Δyと置きかえると近似式になる。Δxが小さいほど、曲線と接線のずれは小さくなる。",
      "entities": [
        "velocity-acceleration-and-approximation"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-vaa-square-approximation",
      "type": "input",
      "prompt": "f(x)＝x²の近似式Δy≈f′(x)Δxを使って、x＝2付近での(2.01)²の値を見積もれ。",
      "choices": null,
      "answer": "4.04",
      "accept": [],
      "explanation": "f′(2)＝4なので、Δy≈4×0.01＝0.04。f(2)＝4に加えると、(2.01)²≈4＋0.04＝4.04。",
      "entities": [
        "velocity-acceleration-and-approximation"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-vaa-velocity-zero-choice",
      "type": "choice",
      "prompt": "位置x(t)＝t³−6t²＋9tで動く点の速度v(t)が0になる時刻t（0＜t）はどれか。",
      "choices": [
        "t＝1、3",
        "t＝1のみ",
        "t＝3のみ",
        "t＝0、2"
      ],
      "answer": 0,
      "accept": [],
      "explanation": "v(t)＝3t²−12t＋9＝3(t−1)(t−3)＝0より、t＝1、3。どちらも点が一瞬止まり、動く向きが変わる時刻。",
      "entities": [
        "velocity-acceleration-and-approximation"
      ],
      "verified": true,
      "placeholder": false
    }
  ]
}