Unit 24 Review Exercises
Differentiation Techniques
These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed.
1 Paper 1 Let \(y = 3x^4 - \dfrac{2}{x} + 5\sqrt{x}\). Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\), and the gradient of the curve at \(x = 1\).
\(y = 3x^4 - 2x^{-1} + 5x^{\frac{1}{2}}\), so \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 12x^3 + 2x^{-2} + \dfrac{5}{2}x^{-\frac{1}{2}} = 12x^3 + \dfrac{2}{x^2} + \dfrac{5}{2\sqrt{x}}\).
At \(x = 1\): \(\;12 + 2 + \dfrac{5}{2} = \dfrac{33}{2}\). See 24.2.
2 Paper 1 Let \(f(x) = (2x - 1)^3(x + 2)^2\). Show that \(f'(x) = 10(2x - 1)^2(x + 2)(x + 1)\).
Product rule with the chain rule:
\[ \begin{aligned} f'(x) &= 3(2x - 1)^2 \cdot 2 \cdot (x + 2)^2 + (2x - 1)^3 \cdot 2(x + 2) \\ &= 2(2x - 1)^2(x + 2)\big[3(x + 2) + (2x - 1)\big] \\ &= 2(2x - 1)^2(x + 2)(5x + 5) \\ &= 10(2x - 1)^2(x + 2)(x + 1) . \end{aligned} \]
See 24.3.
3 Paper 1 Let \(y = \dfrac{x}{x^2 + 9}\). Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\), and the values of \(x\) where the gradient is \(0\).
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(x^2 + 9)(1) - x(2x)}{(x^2 + 9)^2} = \frac{9 - x^2}{(x^2 + 9)^2} . \]
This is \(0\) when \(9 - x^2 = 0\), i.e. \(x = 3\) or \(x = -3\). See 24.3.
4 Paper 1 Let \(f(x) = \mathrm{e}^{2x}\sin x\). Find \(f'(x)\), and hence find \(f'(0)\) and \(f'\left(\dfrac{\pi}{2}\right)\).
5 Paper 1 Let \(f(x) = \ln\left(\dfrac{x^2}{2x + 1}\right)\), \(\;x > 0\). Show that \(f'(x) = \dfrac{2(x + 1)}{x(2x + 1)}\).
Laws of logarithms first: \(\;f(x) = 2\ln x - \ln(2x + 1)\).
\[ f'(x) = \frac{2}{x} - \frac{2}{2x + 1} = \frac{2(2x + 1) - 2x}{x(2x + 1)} = \frac{2x + 2}{x(2x + 1)} = \frac{2(x + 1)}{x(2x + 1)} . \]
See 24.5.
6 Paper 1 Let \(f(x) = \cos^2 3x\). Show that \(f'(x) = -3\sin 6x\), and find \(f'\left(\dfrac{\pi}{12}\right)\).
\(f'(x) = 2\cos 3x \times (-3\sin 3x) = -3(2\sin 3x\cos 3x) = -3\sin 6x\).
\(f'\left(\dfrac{\pi}{12}\right) = -3\sin\dfrac{\pi}{2} = -3\). See 24.4.
7 Paper 1 AHL Let \(y = \arctan 2x + 4^x\). Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and its value at \(x = 0\).
8 Paper 1 AHL The curve \(xy^2 + x^3 = 10\) passes through the point \((1,\,3)\). Find the gradient of the curve at this point.
Check: \(1 \times 9 + 1 = 10\) ✓. Differentiate implicitly:
\[ y^2 + 2xy\frac{\mathrm{d}y}{\mathrm{d}x} + 3x^2 = 0 \quad\Rightarrow\quad \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{y^2 + 3x^2}{2xy} . \]
At \((1,\,3)\): \(\;-\dfrac{9 + 3}{6} = -2\). See 24.6.
9 Paper 1 AHL Let \(y = (2x)^x\), \(\;x > 0\). Use logarithmic differentiation to find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\), and find the gradient of the curve at \(x = \dfrac{1}{2}\).
\(\ln y = x\ln 2x\). Then \(\dfrac{1}{y}\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ln 2x + x \cdot \dfrac{2}{2x} = \ln 2x + 1\), so
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = (2x)^x(\ln 2x + 1) . \]
At \(x = \dfrac{1}{2}\): \(\;1^{\frac{1}{2}}(\ln 1 + 1) = 1\). See 24.6.
10 Paper 2 Let \(f(x) = \dfrac{\mathrm{e}^x}{x^2 + 1}\).
(a) Show that \(f'(x) = \dfrac{\mathrm{e}^x(x - 1)^2}{(x^2 + 1)^2}\), and explain why \(f'(x) \ge 0\) for all \(x\).
(b) Use your GDC to find \(f''(2)\).
(a) Quotient rule:
\[ f'(x) = \frac{(x^2 + 1)\mathrm{e}^x - \mathrm{e}^x(2x)}{(x^2 + 1)^2} = \frac{\mathrm{e}^x(x^2 - 2x + 1)}{(x^2 + 1)^2} = \frac{\mathrm{e}^x(x - 1)^2}{(x^2 + 1)^2} . \]
\(\mathrm{e}^x > 0\), \((x - 1)^2 \ge 0\) and \((x^2 + 1)^2 > 0\), so \(f'(x) \ge 0\).
(b) Menu → Calculus → Numerical Derivative at a Point, value \(2\), 2nd derivative: \(\;f''(2) = 0.414\) (3 s.f.). See 24.3, 24.5 and 24.7.