Derivatives — Calculus I 練習ワークシート

Grades 11–12 / College むずかしい 10問

The derivative of a function measures its instantaneous rate of change — the slope of the tangent line at any point. Notation includes f′(x)f'(x), dydx\frac{dy}{dx}, and ddx[f(x)]\frac{d}{dx}[f(x)].

Core rules: Power rule ddx[xn]=nxn−1\frac{d}{dx}[x^n] = nx^{n-1}; constant rule ddx[c]=0\frac{d}{dx}[c] = 0; sum rule; product rule (fg)′=f′g+fg′(fg)' = f'g + fg'; quotient rule; and chain rule for compositions.

Derivatives appear everywhere in physics (velocity, acceleration), economics (marginal cost), and optimization. Always simplify before differentiating when possible.

練習するスキル

  • Power rule differentiation
  • Product and quotient rules
  • Chain rule for composite functions
  • Finding equations of tangent lines
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練習ワークシート：Derivatives

指示： 各問題を丁寧に解きましょう。途中式もはっきりと書きましょう。最終的な答えは、指定された場所または指示に従って別の用紙に書いてください。

  1. 1.Find the derivative of f(x)=x2sin⁡xexf(x) = \dfrac{x^2 \sin x}{e^x}.

Derivatives — 練習ワークシート（続き）

  1. 2.A particle moves along a line so that its position at time tt (in seconds) is given by s(t)=t3−6t2+9t+2s(t) = t^3 - 6t^2 + 9t + 2 (in meters). Find the velocity of the particle at t=2t = 2 seconds.

Derivatives — 練習ワークシート（続き）

  1. 3.Find dydx\dfrac{dy}{dx} if y=ln⁡(1+cos⁡x1−cos⁡x)y = \ln\left( \sqrt{\dfrac{1 + \cos x}{1 - \cos x}} \right).

Derivatives — 練習ワークシート（続き）

  1. 4.The radius of a sphere is increasing at a rate of 2 cm/s. How fast is the volume increasing when the radius is 5 cm? (Volume of a sphere: V=43πr3V = \frac{4}{3}\pi r^3)
  2. 5.Find the derivative of g(x)=arctan⁡(x1+x2)g(x) = \arctan\left( \dfrac{x}{1 + x^2} \right).

Derivatives — 練習ワークシート（続き）

  1. 6.A ladder 10 feet long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall?
  2. 7.Find f′(x)f'(x) if f(x)=xsin⁡xf(x) = x^{\sin x} for x>0x > 0.

Derivatives — 練習ワークシート（続き）

  1. 8.A conical tank (with vertex down) has a radius of 3 m at the top and a height of 6 m. Water is being pumped into the tank at a rate of 2 m3^3/min. How fast is the water level rising when the water is 4 m deep? (Volume of a cone: V=13πr2hV = \frac{1}{3}\pi r^2 h)

Derivatives — 練習ワークシート（続き）

  1. 9.Find the derivative of h(x)=sec⁡−1(e2x)h(x) = \sec^{-1}(e^{2x}).
  2. 10.A plane flying horizontally at an altitude of 1 mile and a speed of 500 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 2 miles away from the station.

解答集

  1. 1.
    Using quotient rule: f′(x)=(2xsin⁡x+x2cos⁡x)ex−x2sin⁡x⋅exe2x=2xsin⁡x+x2cos⁡x−x2sin⁡xexf'(x) = \frac{(2x\sin x + x^2\cos x)e^x - x^2\sin x \cdot e^x}{e^{2x}} = \frac{2x\sin x + x^2\cos x - x^2\sin x}{e^x}
    最終回答： f′(x)=2xsin⁡x+x2cos⁡x−x2sin⁡xexf'(x) = \dfrac{2x \sin x + x^2 \cos x - x^2 \sin x}{e^x}
  2. 2.
    v(t)=s′(t)=3t2−12t+9v(t) = s'(t) = 3t^2 - 12t + 9. At t=2t=2, v(2)=3(4)−12(2)+9=12−24+9=−3v(2) = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3 m/s (moving downward).
    最終回答： −3-3 m/s
  3. 3.
    Simplify: y=12ln⁡(1+cos⁡x1−cos⁡x)=12[ln⁡(1+cos⁡x)−ln⁡(1−cos⁡x)]y = \frac{1}{2} \ln\left( \frac{1+\cos x}{1-\cos x} \right) = \frac{1}{2} [\ln(1+\cos x) - \ln(1-\cos x)]. Then y′=12(−sin⁡x1+cos⁡x−sin⁡x1−cos⁡x)=12(−sin⁡x(1−cos⁡x)−sin⁡x(1+cos⁡x)1−cos⁡2x)=12(−2sin⁡xsin⁡2x)=−csc⁡xy' = \frac{1}{2} \left( \frac{-\sin x}{1+\cos x} - \frac{\sin x}{1-\cos x} \right) = \frac{1}{2} \left( \frac{-\sin x(1-\cos x) - \sin x(1+\cos x)}{1-\cos^2 x} \right) = \frac{1}{2} \left( \frac{-2\sin x}{\sin^2 x} \right) = -\csc x.
    最終回答： −csc⁡x-\csc x
  4. 4.
    dVdt=4πr2drdt\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}. Given drdt=2\frac{dr}{dt} = 2 cm/s, r=5r=5: dVdt=4π(25)(2)=200π\frac{dV}{dt} = 4\pi (25)(2) = 200\pi cm3^3/s.
    最終回答： 200π200\pi cm3^3/s

解答集（続き）

  1. 5.
    g′(x)=11+(x1+x2)2⋅(1)(1+x2)−x(2x)(1+x2)2=11+x2(1+x2)2⋅1+x2−2x2(1+x2)2=1(1+x2)2+x2(1+x2)2⋅1−x2(1+x2)2=(1+x2)2(1+x2)2+x2⋅1−x2(1+x2)2=1−x2(1+x2)2+x2g'(x) = \frac{1}{1 + \left( \frac{x}{1+x^2} \right)^2} \cdot \frac{(1)(1+x^2) - x(2x)}{(1+x^2)^2} = \frac{1}{1 + \frac{x^2}{(1+x^2)^2}} \cdot \frac{1+x^2 - 2x^2}{(1+x^2)^2} = \frac{1}{\frac{(1+x^2)^2 + x^2}{(1+x^2)^2}} \cdot \frac{1 - x^2}{(1+x^2)^2} = \frac{(1+x^2)^2}{(1+x^2)^2 + x^2} \cdot \frac{1 - x^2}{(1+x^2)^2} = \frac{1 - x^2}{(1+x^2)^2 + x^2}.
    最終回答： g′(x)=1−x2(1+x2)2+x2g'(x) = \dfrac{1 - x^2}{(1+x^2)^2 + x^2}
  2. 6.
    Let xx be distance from wall to bottom, yy be height of top on wall. x2+y2=100x^2 + y^2 = 100. Differentiate: 2xdxdt+2ydydt=0⇒dydt=−xydxdt2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}. When x=6x=6, y=8y=8, dxdt=1\frac{dx}{dt}=1: dydt=−68(1)=−34\frac{dy}{dt} = -\frac{6}{8}(1) = -\frac{3}{4} ft/s (sliding down at 0.75 ft/s).
    最終回答： −34-\frac{3}{4} ft/s
  3. 7.
    f(x)=esin⁡xln⁡xf(x) = e^{\sin x \ln x}. Then f′(x)=esin⁡xln⁡x(cos⁡xln⁡x+sin⁡xx)=xsin⁡x(cos⁡xln⁡x+sin⁡xx)f'(x) = e^{\sin x \ln x} \left( \cos x \ln x + \frac{\sin x}{x} \right) = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right).
    最終回答： f′(x)=xsin⁡x(cos⁡xln⁡x+sin⁡xx)f'(x) = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right)

解答集（続き）

  1. 8.
    By similar triangles, rh=36=12\frac{r}{h} = \frac{3}{6} = \frac{1}{2}, so r=h2r = \frac{h}{2}. Then V=13π(h2)2h=π12h3V = \frac{1}{3}\pi \left( \frac{h}{2} \right)^2 h = \frac{\pi}{12} h^3. Differentiate: dVdt=π4h2dhdt\frac{dV}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}. Given dVdt=2\frac{dV}{dt}=2, h=4h=4: 2=π4(16)dhdt⇒dhdt=24π=12π2 = \frac{\pi}{4}(16) \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{2}{4\pi} = \frac{1}{2\pi} m/min.
    最終回答： 12π\frac{1}{2\pi} m/min
  2. 9.
    h′(x)=1∣e2x∣(e2x)2−1⋅2e2x=2e2xe2xe4x−1=2e4x−1h'(x) = \frac{1}{|e^{2x}| \sqrt{(e^{2x})^2 - 1}} \cdot 2e^{2x} = \frac{2e^{2x}}{e^{2x} \sqrt{e^{4x} - 1}} = \frac{2}{\sqrt{e^{4x} - 1}}.
    最終回答： h′(x)=2e4x−1h'(x) = \dfrac{2}{\sqrt{e^{4x} - 1}}
  3. 10.
    Let xx be horizontal distance from station to plane, ss be distance from station to plane. s2=x2+12s^2 = x^2 + 1^2. Differentiate: 2sdsdt=2xdxdt⇒dsdt=xsdxdt2s \frac{ds}{dt} = 2x \frac{dx}{dt} \Rightarrow \frac{ds}{dt} = \frac{x}{s} \frac{dx}{dt}. When s=2s=2, x=4−1=3x = \sqrt{4-1} = \sqrt{3}, dxdt=500\frac{dx}{dt}=500: dsdt=32(500)=2503\frac{ds}{dt} = \frac{\sqrt{3}}{2}(500) = 250\sqrt{3} mi/h.
    最終回答： 2503250\sqrt{3} mi/h
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避けるべきよくある間違い

  • Forgetting the chain rule on nested functions
  • Product rule errors (differentiating only one factor)
  • Leaving negative exponents unconverted before applying power rule
  • Confusing average rate of change with instantaneous derivative

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