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IB Mathematics: Applications and Interpretation HL · AHL 2: Further Derivatives · AHL 2.4

The product and quotient rules Problem set B Worked solutions

Skills

  1. Differentiate with respect to x:

    1. (a) \(y=x^7\cos x\)

      • \(u=x^7,\; v=\cos x\); \(u'=7x^6,\; v'=-\sin x\)
      • \(\dfrac{dy}{dx}=7x^6\cos x-x^7\sin x=x^6(7\cos x-x\sin x)\)

    2. (b) \(y=x^{5/2}e^x\)

      • \(u=x^{5/2},\; v=e^x\); \(u'=\tfrac52x^{3/2},\; v'=e^x\)
      • \(\dfrac{dy}{dx}=\tfrac52x^{3/2}e^x+x^{5/2}e^x\)

    3. (c) \(y=x^8\ln x\)

      • \(u=x^8,\; v=\ln x\); \(u'=8x^7,\; v'=\dfrac1x\)
      • \(\dfrac{dy}{dx}=8x^7\ln x+x^8\cdot\dfrac1x=x^7(8\ln x+1)\)

  2. Differentiate with respect to x:

    1. (a) \(f(x)=x^2\cos5x\)

      • \(u=x^2,\; v=\cos5x\); \(u'=2x,\; v'=-5\sin5x\)
      • \(f'(x)=2x\cos5x-5x^2\sin5x\)

    2. (b) \(f(x)=x^3e^{3x+4}\)

      • \(u=x^3,\; v=e^{3x+4}\); \(u'=3x^2,\; v'=3e^{3x+4}\)
      • \(f'(x)=3x^2e^{3x+4}+3x^3e^{3x+4}=3x^2e^{3x+4}(1+x)\)

    3. (c) \(f(x)=x^2\ln(4x+1)\)

      • \(u=x^2,\; v=\ln(4x+1)\); \(u'=2x,\; v'=\dfrac{4}{4x+1}\)
      • \(f'(x)=2x\ln(4x+1)+\dfrac{4x^2}{4x+1}\)

  3. Use the quotient rule to find \(\dfrac{dy}{dx}\). Simplify your answers.

    1. (a) \(y=\dfrac{4x-3}{x^2+5}\)

      • \(u=4x-3,\; v=x^2+5\); \(u'=4,\; v'=2x\)
      • \(\dfrac{dy}{dx}=\dfrac{4(x^2+5)-(4x-3)(2x)}{(x^2+5)^2}=\dfrac{-4x^2+6x+20}{(x^2+5)^2}\)

    2. (b) \(y=\dfrac{x-5}{\sqrt{2x+1}}\)

      • \(u=x-5,\; v=(2x+1)^{1/2}\); \(u'=1,\; v'=(2x+1)^{-1/2}\)
      • \(\dfrac{dy}{dx}=\dfrac{(2x+1)^{1/2}-(x-5)(2x+1)^{-1/2}}{2x+1}\)
      • Multiply top and bottom by \((2x+1)^{1/2}\): \(\dfrac{(2x+1)-(x-5)}{(2x+1)^{3/2}}=\dfrac{x+6}{(2x+1)^{3/2}}\)

    3. (c) \(y=\dfrac{e^x}{(5x+2)^2}\)

      • \(u=e^x,\; v=(5x+2)^2\); \(u'=e^x,\; v'=10(5x+2)\)
      • \(\dfrac{dy}{dx}=\dfrac{e^x(5x+2)^2-10e^x(5x+2)}{(5x+2)^4}\)
      • Cancel \((5x+2)\): \(\dfrac{e^x(5x+2-10)}{(5x+2)^3}=\dfrac{e^x(5x-8)}{(5x+2)^3}\)

    4. (d) \(y=\dfrac{e^x}{(2-x)^4}\)

      • \(u=e^x,\; v=(2-x)^4\); \(u'=e^x,\; v'=-4(2-x)^3\)
      • \(\dfrac{dy}{dx}=\dfrac{e^x(2-x)^4+4e^x(2-x)^3}{(2-x)^8}\)
      • Cancel \((2-x)^3\): \(\dfrac{e^x(2-x+4)}{(2-x)^5}=\dfrac{e^x(6-x)}{(2-x)^5}\)

  4. Differentiate, and where asked evaluate to 3 significant figures.

    1. (a) If \(f(x)=xe^{4x}\), find \(f'(0.5)\).

      • \(f'(x)=e^{4x}+4xe^{4x}=e^{4x}(1+4x)\)
      • From the calculator: \(f'(0.5)=3e^2=22.2\)

    2. (b) \(y=x^3\sin x\ln x\)

      • Three factors: differentiate one at a time, keeping the other two. \(u=x^3,\; v=\sin x,\; w=\ln x\)
      • \(\dfrac{dy}{dx}=3x^2\sin x\ln x+x^3\cos x\ln x+x^3\sin x\cdot\dfrac1x\)
      • \(=3x^2\sin x\ln x+x^3\cos x\ln x+x^2\sin x\)

    3. (c) \(y=e^{x\sin2x}\)

      • Chain rule: the inside is \(x\sin2x\), a product. Its derivative is \(\sin2x+2x\cos2x\)
      • \(\dfrac{dy}{dx}=e^{x\sin2x}(\sin2x+2x\cos2x)\)

  5. Give exact answers.

    1. (a) Find the equation of the tangent to \(y=\dfrac{x}{x^2+3}\) at \(x=1\).

      • When \(x=1\): \(y=\dfrac{1}{4}\)
      • \(\dfrac{dy}{dx}=\dfrac{1(x^2+3)-x(2x)}{(x^2+3)^2}=\dfrac{3-x^2}{(x^2+3)^2}\); at \(x=1\) the gradient is \(\dfrac{2}{16}=\dfrac18\)
      • \(y-\dfrac14=\dfrac18(x-1)\), so \(y=\dfrac{x+1}{8}\)

    2. (b) Find the equation of the normal to \(y=x\cos2x\) at \(x=\dfrac{\pi}{4}\).

      • When \(x=\dfrac{\pi}{4}\): \(y=\dfrac{\pi}{4}\cos\dfrac{\pi}{2}=0\)
      • \(\dfrac{dy}{dx}=\cos2x-2x\sin2x\); at \(x=\dfrac{\pi}{4}\) the gradient is \(0-\dfrac{\pi}{2}=-\dfrac{\pi}{2}\)
      • Normal gradient \(=\dfrac{2}{\pi}\): \(y-0=\dfrac{2}{\pi}\left(x-\dfrac{\pi}{4}\right)\), so \(y=\dfrac{2x}{\pi}-\dfrac12\)

  6. Find the coordinates of the point where the gradient of the curve is as given. Give coordinates to 3 significant figures.

    1. (a) \(y=e^{3x}-4x\), where the gradient is 5.

      • \(\dfrac{dy}{dx}=3e^{3x}-4\)
      • \(3e^{3x}-4=5\Rightarrow e^{3x}=3\Rightarrow x=\dfrac{\ln3}{3}\)
      • \(y=e^{\ln3}-4\cdot\dfrac{\ln3}{3}=3-\dfrac{4\ln3}{3}=1.54\)

    2. (b) \(y=x\ln3x\), where the gradient is 4.

      • \(\dfrac{dy}{dx}=\ln3x+x\cdot\dfrac{1}{x}=\ln3x+1\)
      • \(\ln3x+1=4\Rightarrow 3x=e^3\Rightarrow x=\dfrac{e^3}{3}\)
      • \(y=\dfrac{e^3}{3}\ln e^3=e^3=20.1\)

In context

  1. The profit, in thousands of pounds, from a new app is modelled by \(P=8te^{-0.25t}\), where t is the time in months after the launch, \(t\ge0\).

    1. (a) Find \(\dfrac{dP}{dt}\).

      • \(u=8t,\; v=e^{-0.25t}\); \(u'=8,\; v'=-0.25e^{-0.25t}\)
      • \(\dfrac{dP}{dt}=8e^{-0.25t}-2te^{-0.25t}=8e^{-0.25t}(1-0.25t)\)

    2. (b) Find when the profit is greatest, and its value then.

      • \(\dfrac{dP}{dt}=0\): \(e^{-0.25t}\) is never 0, so \(1-0.25t=0\) and \(t=4\)
      • \(P=8(4)e^{-1}=11.8\) thousand pounds

    3. (c) Find the rate of change of the profit when \(t=8\), with its units, and say what it tells you.

      • \(\dfrac{dP}{dt}=8e^{-2}(1-2)\)
      • From the calculator: \(-1.08\). The negative sign means the profit is falling.

  2. The number of visitors, in thousands, on a website t days after it opens is \(V=\dfrac{30t}{t^2+25}\), \(t\ge0\).

    1. (a) Find \(\dfrac{dV}{dt}\), and simplify.

      • \(u=30t,\; v=t^2+25\); \(u'=30,\; v'=2t\)
      • \(\dfrac{dV}{dt}=\dfrac{30(t^2+25)-30t(2t)}{(t^2+25)^2}=\dfrac{30(25-t^2)}{(t^2+25)^2}\)

    2. (b) Find the rate of change of the number of visitors when \(t=3\), with its units.

      • \(\dfrac{dV}{dt}=\dfrac{30(25-9)}{(9+25)^2}=\dfrac{480}{1156}\)
      • From the calculator: 0.415

    3. (c) Find when the number of visitors is greatest, and its value then.

      • \(\dfrac{dV}{dt}=0\Rightarrow 25-t^2=0\Rightarrow t=5\) (t is positive)
      • \(V=\dfrac{30(5)}{50}=3\)

Stretch

  1. Differentiate and simplify.

    1. (a) If \(f(x)=x\sqrt{1+3x}\), show that \(f'(x)=\dfrac{ax+b}{2\sqrt{1+3x}}\), where a and b are integers to be found.

      • \(u=x,\; v=(1+3x)^{1/2}\); \(u'=1,\; v'=\tfrac32(1+3x)^{-1/2}\)
      • \(f'(x)=(1+3x)^{1/2}+\dfrac{3x}{2(1+3x)^{1/2}}\)
      • Common denominator: \(f'(x)=\dfrac{2(1+3x)+3x}{2\sqrt{1+3x}}=\dfrac{9x+2}{2\sqrt{1+3x}}\)

    2. (b) Find and simplify an expression for the derivative of \(f(x)=\dfrac{x}{\sqrt{k+2x^2}}\), where k is a positive constant.

      • \(u=x,\; v=(k+2x^2)^{1/2}\); \(u'=1,\; v'=2x(k+2x^2)^{-1/2}\)
      • \(f'(x)=\dfrac{(k+2x^2)^{1/2}-2x^2(k+2x^2)^{-1/2}}{k+2x^2}\)
      • Multiply top and bottom by \((k+2x^2)^{1/2}\): \(\dfrac{(k+2x^2)-2x^2}{(k+2x^2)^{3/2}}=\dfrac{k}{(k+2x^2)^{3/2}}\)

  2. Show that \(f(x)=\dfrac{x^2}{x-4}\) is an increasing function for \(x>a\) and for \(x<b\), where a and b are constants to be found.

    • \(f'(x)=\dfrac{2x(x-4)-x^2}{(x-4)^2}=\dfrac{x^2-8x}{(x-4)^2}=\dfrac{x(x-8)}{(x-4)^2}\)
    • \((x-4)^2>0\), so the sign of \(f'\) is the sign of \(x(x-8)\)
    • \(f'(x)>0\) when \(x>8\) or \(x<0\). So f is increasing there: \(a=8\), \(b=0\)

  3. Use the quotient rule to show that if \(f(x)=\sec x\), where \(\sec x=\dfrac{1}{\cos x}\), then \(f'(x)=\dfrac{\sin x}{\cos^2x}\).

    • \(u=1,\; v=\cos x\); \(u'=0,\; v'=-\sin x\)
    • \(f'(x)=\dfrac{0\cdot\cos x-1\cdot(-\sin x)}{\cos^2x}\)
    • \(f'(x)=\dfrac{\sin x}{\cos^2x}\)

  4. Let \(y=2\sin^2x\), for \(0<x<2\pi\).

    1. (a) Find \(\dfrac{dy}{dx}\).

      • Chain rule on \(2(\sin x)^2\): \(2\times2(\sin x)\times\cos x\)
      • \(\dfrac{dy}{dx}=4\sin x\cos x\)

    2. (b) Find the values of x where the gradient of the curve is zero.

      • \(4\sin x\cos x=0\Rightarrow\sin x=0\) or \(\cos x=0\)
      • In \(0<x<2\pi\): \(x=\pi\) from \(\sin x=0\); \(x=\dfrac{\pi}{2},\dfrac{3\pi}{2}\) from \(\cos x=0\)

    3. (c) Given that \(\cos2x=2\cos^2x-1\), differentiate both sides to show that \(\sin2x=2\sin x\cos x\).

      • Differentiate \(\cos2x=2\cos^2x-1\): \(-2\sin2x=-4\cos x\sin x\)
      • Divide by \(-2\): \(\sin2x=2\sin x\cos x\)

    4. (d) Given that \(\sin4x=2\sin2x\cos2x\), use differentiation to show that \(\cos4x=\cos^22x-\sin^22x\).

      • Differentiate \(\sin4x=2\sin2x\cos2x\): \(4\cos4x=2\left[2\cos2x\cos2x+\sin2x(-2\sin2x)\right]\)
      • \(4\cos4x=4\cos^22x-4\sin^22x\)
      • \(\cos4x=\cos^22x-\sin^22x\)

  5. Let \(y=e^{4x}\).

    1. (a) Show that \(\dfrac{dy}{dx}=4y\).

      • \(\dfrac{dy}{dx}=4e^{4x}\)
      • Since \(y=e^{4x}\), \(\dfrac{dy}{dx}=4y\)

    2. (b) Is it true that if \(\dfrac{dy}{dx}=4y\), then \(y=e^{4x}\)? Justify your answer.

      • Try \(y=7e^{4x}\): \(\dfrac{dy}{dx}=28e^{4x}=4(7e^{4x})=4y\)
      • So \(dy/dx=4y\) does not force \(y=e^{4x}\). Any constant multiple works

    3. (c) Now let \(y=e^{kx}\), where \(k>0\). The tangent to this curve at \(x=\dfrac3k\) is L. Prove that the y-intercept of L is independent of k.

      • At \(x=\dfrac3k\): \(y=e^3\) and \(\dfrac{dy}{dx}=ke^3\)
      • Tangent: \(y-e^3=ke^3\left(x-\dfrac3k\right)\)
      • \(y=ke^3x-3e^3+e^3=ke^3x-2e^3\). At \(x=0\), \(y=-2e^3\), the same for every k

  6. Find the x-coordinate of every point on the curve \(y=\ln(x^2-6)\) where the tangent is parallel to the line \(y=3x-1\).

    • \(\dfrac{dy}{dx}=\dfrac{2x}{x^2-6}\). Parallel to \(y=3x-1\) means gradient 3
    • \(\dfrac{2x}{x^2-6}=3\Rightarrow 3x^2-2x-18=0\)
    • From the calculator: \(x=2.81\) or \(x=-2.14\)
    • The curve exists only where \(x^2>6\). \((-2.14)^2=4.58\) fails, \(2.81^2=7.9\) works, so \(x=2.81\)

jmaths.xyz · AI HL · AHL 2.4 The product and quotient rules