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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 Answers
Skills
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- (a) \(x^4(5\sin x+x\cos x)\)
- (b) \(e^x\left(x^{3/2}+\tfrac32x^{1/2}\right)\)
- (c) \(x^5(6\ln x+1)\)
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- (a) \(3x^2\sin 4x+4x^3\cos 4x\)
- (b) \(xe^{5x-2}(2+5x)\)
- (c) \(3x^2\ln(2x+5)+\dfrac{2x^3}{2x+5}\)
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- (a) \(\dfrac{-3x^2+2x+6}{(x^2+2)^2}\)
- (b) \(\dfrac{3x-16}{2(3x-2)^{3/2}}\)
- (c) \(\dfrac{e^x(4x-7)}{(4x+1)^3}\)
- (d) \(\dfrac{e^x(6-x)}{(3-x)^4}\)
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- (a) 2820
- (b) \(xe^x(2\ln x+x\ln x+1)\)
- (c) \(e^{x\cos x}(\cos x-x\sin x)\)
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- (a) \(y=-\dfrac{3}{25}x+\dfrac{16}{25}\)
- (b) \(y=-x+\dfrac{\pi}{3}\)
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- (a) \((0.693,\;0.534)\)
- (b) \((3.69,\;7.39)\)
In context
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- (a) \(\dfrac{dC}{dt}=12e^{-0.5t}(1-0.5t)\)
- (b) t = 2 hours; C = 8.83 mg/l
- (c) −1.62 mg/l per hour, so the concentration is falling
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- (a) \(\dfrac{dC}{dt}=\dfrac{20(16-t^2)}{(t^2+16)^2}\)
- (b) 0.6 mg/l per day
- (c) t = 4 days; C = 2.5 mg/l
Stretch
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- (a) \(a=-3,\; b=10\)
- (b) \(\dfrac{k}{(k-x^2)^{3/2}}\)
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\(a=6,\; b=0\)
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\(f'(x)=-\dfrac{1}{\sin^2x}\)
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- (a) \(\dfrac{dy}{dx}=-2\sin x\cos x\)
- (b) \(x=\dfrac{\pi}{2},\; \pi,\; \dfrac{3\pi}{2}\)
- (c) \(\sin2x=2\sin x\cos x\)
- (d) \(\sin3x=3\sin x-4\sin^3x\)
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- (a) \(\dfrac{dy}{dx}=3e^{3x}=3y\)
- (b) No. For example \(y=2e^{3x}\) also has \(\dfrac{dy}{dx}=3y\), and it is not \(e^{3x}\)
- (c) y-intercept \(=-e^2\), which does not contain k
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\(x=2.5\)
jmaths.xyz · AI HL · AHL 2.4 The product and quotient rules