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

The product and quotient rules Problem set A One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Skills

  1. Differentiate with respect to x:

    1. (a) \(y=x^5\sin x\)

      • \(u=x^5,\; v=\sin x\); \(u'=5x^4,\; v'=\cos x\)
      • \(\dfrac{dy}{dx}=5x^4\sin x+x^5\cos x=x^4(5\sin x+x\cos x)\)

    2. (b) \(y=x^{3/2}e^x\)

      • \(u=x^{3/2},\; v=e^x\); \(u'=\tfrac32x^{1/2},\; v'=e^x\)
      • \(\dfrac{dy}{dx}=\tfrac32x^{1/2}e^x+x^{3/2}e^x\)

    3. (c) \(y=x^6\ln x\)

      • \(u=x^6,\; v=\ln x\); \(u'=6x^5,\; v'=\dfrac1x\)
      • \(\dfrac{dy}{dx}=6x^5\ln x+x^6\cdot\dfrac1x=x^5(6\ln x+1)\)

  2. Differentiate with respect to x:

    1. (a) \(f(x)=x^3\sin4x\)

      • \(u=x^3,\; v=\sin4x\); \(u'=3x^2,\; v'=4\cos4x\)
      • \(f'(x)=3x^2\sin4x+4x^3\cos4x\)

    2. (b) \(f(x)=x^2e^{5x-2}\)

      • \(u=x^2,\; v=e^{5x-2}\); \(u'=2x,\; v'=5e^{5x-2}\)
      • \(f'(x)=2xe^{5x-2}+5x^2e^{5x-2}=xe^{5x-2}(2+5x)\)

    3. (c) \(f(x)=x^3\ln(2x+5)\)

      • \(u=x^3,\; v=\ln(2x+5)\); \(u'=3x^2,\; v'=\dfrac{2}{2x+5}\)
      • \(f'(x)=3x^2\ln(2x+5)+\dfrac{2x^3}{2x+5}\)

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

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

      • \(u=3x-1,\; v=x^2+2\); \(u'=3,\; v'=2x\)
      • \(\dfrac{dy}{dx}=\dfrac{3(x^2+2)-(3x-1)(2x)}{(x^2+2)^2}=\dfrac{-3x^2+2x+6}{(x^2+2)^2}\)

    2. (b) \(y=\dfrac{x+4}{\sqrt{3x-2}}\)

      • \(u=x+4,\; v=(3x-2)^{1/2}\); \(u'=1,\; v'=\tfrac32(3x-2)^{-1/2}\)
      • \(\dfrac{dy}{dx}=\dfrac{(3x-2)^{1/2}-\tfrac32(x+4)(3x-2)^{-1/2}}{3x-2}\)
      • Multiply top and bottom by \((3x-2)^{1/2}\): \(\dfrac{(3x-2)-\tfrac32(x+4)}{(3x-2)^{3/2}}=\dfrac{3x-16}{2(3x-2)^{3/2}}\)

    3. (c) \(y=\dfrac{e^x}{(4x+1)^2}\)

      • \(u=e^x,\; v=(4x+1)^2\); \(u'=e^x,\; v'=8(4x+1)\)
      • \(\dfrac{dy}{dx}=\dfrac{e^x(4x+1)^2-8e^x(4x+1)}{(4x+1)^4}\)
      • Cancel \((4x+1)\): \(\dfrac{e^x(4x+1-8)}{(4x+1)^3}=\dfrac{e^x(4x-7)}{(4x+1)^3}\)

    4. (d) \(y=\dfrac{e^x}{(3-x)^3}\)

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

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

    1. (a) If \(f(x)=xe^{3x}\), find \(f'(2)\).

      • \(f'(x)=e^{3x}+3xe^{3x}=e^{3x}(1+3x)\)
      • From the calculator: \(f'(2)=7e^6=2820\)

    2. (b) \(y=x^2e^x\ln x\)

      • Three factors: differentiate one at a time, keeping the other two. \(u=x^2,\; v=e^x,\; w=\ln x\)
      • \(\dfrac{dy}{dx}=2xe^x\ln x+x^2e^x\ln x+x^2e^x\cdot\dfrac1x\)
      • \(=xe^x(2\ln x+x\ln x+1)\)

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

      • Chain rule: the inside is \(x\cos x\), a product. Its derivative is \(\cos x-x\sin x\)
      • \(\dfrac{dy}{dx}=e^{x\cos x}(\cos x-x\sin x)\)

  5. Give exact answers.

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

      • When \(x=2\): \(y=\dfrac{2}{5}\)
      • \(\dfrac{dy}{dx}=\dfrac{1(x^2+1)-x(2x)}{(x^2+1)^2}=\dfrac{1-x^2}{(x^2+1)^2}\); at \(x=2\) the gradient is \(-\dfrac{3}{25}\)
      • \(y-\dfrac25=-\dfrac{3}{25}(x-2)\), so \(y=-\dfrac{3}{25}x+\dfrac{16}{25}\)

    2. (b) Find the equation of the normal to \(y=x\sin3x\) at \(x=\dfrac{\pi}{6}\).

      • When \(x=\dfrac{\pi}{6}\): \(y=\dfrac{\pi}{6}\sin\dfrac{\pi}{2}=\dfrac{\pi}{6}\)
      • \(\dfrac{dy}{dx}=\sin3x+3x\cos3x\); at \(x=\dfrac{\pi}{6}\) the gradient is \(1+0=1\)
      • Normal gradient \(=-1\): \(y-\dfrac{\pi}{6}=-\left(x-\dfrac{\pi}{6}\right)\), so \(y=-x+\dfrac{\pi}{3}\)

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

    1. (a) \(y=e^{2x}-5x\), where the gradient is 3.

      • \(\dfrac{dy}{dx}=2e^{2x}-5\)
      • \(2e^{2x}-5=3\Rightarrow e^{2x}=4\Rightarrow x=\tfrac12\ln4=\ln2\)
      • \(y=e^{2\ln2}-5\ln2=4-5\ln2=0.534\)

    2. (b) \(y=x\ln2x\), where the gradient is 3.

      • \(\dfrac{dy}{dx}=\ln2x+x\cdot\dfrac{1}{x}=\ln2x+1\)
      • \(\ln2x+1=3\Rightarrow 2x=e^2\Rightarrow x=\dfrac{e^2}{2}\)
      • \(y=\dfrac{e^2}{2}\ln e^2=e^2=7.39\)

In context

  1. The concentration of a drug in the blood, in mg/l, is modelled by \(C=12te^{-0.5t}\), where t is the time in hours after an injection, \(t\ge0\).

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

      • \(u=12t,\; v=e^{-0.5t}\); \(u'=12,\; v'=-0.5e^{-0.5t}\)
      • \(\dfrac{dC}{dt}=12e^{-0.5t}-6te^{-0.5t}=12e^{-0.5t}(1-0.5t)\)

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

      • \(\dfrac{dC}{dt}=0\): \(e^{-0.5t}\) is never 0, so \(1-0.5t=0\) and \(t=2\)
      • \(C=12(2)e^{-1}=8.83\) mg/l

    3. (c) Find the rate of change of the concentration when \(t=4\), with its units, and say what it tells you.

      • \(\dfrac{dC}{dt}=12e^{-2}(1-2)\)
      • From the calculator: \(-1.62\). The negative sign means the concentration is falling.

  2. The concentration of a pollutant in a lake, in mg/l, t days after a spill is \(C=\dfrac{20t}{t^2+16}\), \(t\ge0\).

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

      • \(u=20t,\; v=t^2+16\); \(u'=20,\; v'=2t\)
      • \(\dfrac{dC}{dt}=\dfrac{20(t^2+16)-20t(2t)}{(t^2+16)^2}=\dfrac{20(16-t^2)}{(t^2+16)^2}\)

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

      • \(\dfrac{dC}{dt}=\dfrac{20(16-4)}{(4+16)^2}=\dfrac{240}{400}=0.6\)

    3. (c) Find when the concentration is greatest, and its value then.

      • \(\dfrac{dC}{dt}=0\Rightarrow 16-t^2=0\Rightarrow t=4\) (t is positive)
      • \(C=\dfrac{20(4)}{32}=2.5\)

Stretch

  1. Differentiate and simplify.

    1. (a) If \(f(x)=x\sqrt{5-x}\), show that \(f'(x)=\dfrac{ax+b}{2\sqrt{5-x}}\), where a and b are integers to be found.

      • \(u=x,\; v=(5-x)^{1/2}\); \(u'=1,\; v'=-\tfrac12(5-x)^{-1/2}\)
      • \(f'(x)=(5-x)^{1/2}-\dfrac{x}{2(5-x)^{1/2}}\)
      • Common denominator: \(f'(x)=\dfrac{2(5-x)-x}{2\sqrt{5-x}}=\dfrac{10-3x}{2\sqrt{5-x}}\)

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

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

  2. Show that \(f(x)=\dfrac{x^2}{x-3}\) is an increasing function for \(x>a\) and for \(x<b\), where a and b are constants to be found.

    • \(f'(x)=\dfrac{2x(x-3)-x^2}{(x-3)^2}=\dfrac{x^2-6x}{(x-3)^2}=\dfrac{x(x-6)}{(x-3)^2}\)
    • \((x-3)^2>0\), so the sign of \(f'\) is the sign of \(x(x-6)\)
    • \(f'(x)>0\) when \(x>6\) or \(x<0\). So f is increasing there: \(a=6\), \(b=0\)

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

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

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

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

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

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

      • \(-2\sin x\cos x=0\Rightarrow\sin x=0\) 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 \(\cos2x=1-2\sin^2x\), differentiate both sides to show that \(\sin2x=2\sin x\cos x\).

      • Differentiate \(\cos2x=1-2\sin^2x\): \(-2\sin2x=-2\times2\sin x\cos x\)
      • Divide by \(-2\): \(\sin2x=2\sin x\cos x\)

    4. (d) Given that \(\cos3x=4\cos^3x-3\cos x\), use differentiation to show that \(\sin3x=3\sin x-4\sin^3x\).

      • Differentiate \(\cos3x=4\cos^3x-3\cos x\): \(-3\sin3x=-12\cos^2x\sin x+3\sin x\)
      • \(\sin3x=4\cos^2x\sin x-\sin x=4\sin x(1-\sin^2x)-\sin x\)
      • \(\sin3x=3\sin x-4\sin^3x\)

  5. Let \(y=e^{3x}\).

    1. (a) Show that \(\dfrac{dy}{dx}=3y\).

      • \(\dfrac{dy}{dx}=3e^{3x}\)
      • Since \(y=e^{3x}\), \(\dfrac{dy}{dx}=3y\)

    2. (b) Is it true that if \(\dfrac{dy}{dx}=3y\), then \(y=e^{3x}\)? Justify your answer.

      • Try \(y=2e^{3x}\): \(\dfrac{dy}{dx}=6e^{3x}=3(2e^{3x})=3y\)
      • So \(dy/dx=3y\) does not force \(y=e^{3x}\). Any constant multiple works

    3. (c) Now let \(y=e^{kx}\), where \(k>0\). The tangent to this curve at \(x=\dfrac2k\) is L. Prove that the y-intercept of L is independent of k.

      • At \(x=\dfrac2k\): \(y=e^2\) and \(\dfrac{dy}{dx}=ke^2\)
      • Tangent: \(y-e^2=ke^2\left(x-\dfrac2k\right)\)
      • \(y=ke^2x-2e^2+e^2=ke^2x-e^2\). At \(x=0\), \(y=-e^2\), the same for every k

  6. Find the x-coordinate of every point on the curve \(y=\ln(x^2-5)\) where the tangent is parallel to the line \(y=4x+3\).

    • \(\dfrac{dy}{dx}=\dfrac{2x}{x^2-5}\). Parallel to \(y=4x+3\) means gradient 4
    • \(\dfrac{2x}{x^2-5}=4\Rightarrow 4x^2-2x-20=0\Rightarrow 2x^2-x-10=0\)
    • \((2x-5)(x+2)=0\): \(x=2.5\) or \(x=-2\)
    • The curve exists only where \(x^2>5\). \((-2)^2=4\) fails, \(2.5^2=6.25\) works, so \(x=2.5\)

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