P7.1First principles, second derivatives and concavity

• M1 for forming the gradient of a chord, \(\frac{f(x + h) - f(x)}{h} = \frac{4(x + h)^{2} - 6(x + h) - 8 - (4x^{2} - 6x - 8)}{h}\)
• A1 for a correct numerator, \(8xh + 4h^{2} - 6h\)
• A1 for dividing by \(h\) to obtain \(8x - 6 + 4h\)
• A1* for a complete proof: as \(h \to 0\), \(8x - 6 + 4h \to 8x - 6\), so the derivative is \(8x - 6\) (cso)

Worked solution: Let \(f(x) = 4x^{2} - 6x - 8\).
\(\frac{f(x + h) - f(x)}{h} = \frac{4(x + h)^{2} - 6(x + h) - 8 - (4x^{2} - 6x - 8)}{h} = \frac{8xh + 4h^{2} - 6h}{h} = 8x - 6 + 4h\)
So \(f'(x) = \lim_{h \to 0}\left(8x - 6 + 4h\right) = 8x - 6\).

• B1* for using \(\sin(x + h) = \sin x\cos h + \cos x\sin h\) and rearranging

Worked solution: \(\sin(x + h) - \sin x = \sin x\cos h + \cos x\sin h - \sin x = \sin x(\cos h - 1) + \cos x\sin h\).

• M1 for the gradient of the chord: \(\frac{\sin(x + h) - \sin x}{h}\) and the limit as \(h \to 0\)
• M1 for using part (a) and splitting: \(\sin x\left(\frac{\cos h - 1}{h}\right) + \cos x\left(\frac{\sin h}{h}\right)\)
• M1 for applying both given limits
• A1* for \(\frac{\mathrm{d}}{\mathrm{d}x}(\sin x) = \cos x\), with correct limit notation

Worked solution: \(\dfrac{\mathrm{d}}{\mathrm{d}x}(\sin x) = \lim_{h \to 0} \dfrac{\sin(x + h) - \sin x}{h} = \lim_{h \to 0}\left[\sin x\left(\dfrac{\cos h - 1}{h}\right) + \cos x\left(\dfrac{\sin h}{h}\right)\right]\).
As \(h \to 0\), \(\dfrac{\cos h - 1}{h} \to 0\) and \(\dfrac{\sin h}{h} \to 1\), so the limit is \(\sin x \times 0 + \cos x \times 1 = \cos x\).

• B1 for the given limits (small angle approximations) hold only for angles in radians

Worked solution: The result \(\frac{\sin h}{h} \to 1\) comes from \(\sin h \approx h\) for small \(h\), which is true only when \(h\) is measured in radians. (In degrees the derivative would be \(\frac{\pi}{180}\cos x\).)

• B1 for a correct expansion, \((x + h)^{4} = x^{4} + 4x^{3}h + 6x^{2}h^{2} + 4xh^{3} + h^{4}\), seen or implied
• M1 for forming the gradient of a chord, \(\frac{f(x + h) - f(x)}{h} = \frac{2(x + h)^{4} - 5(x + h) - (2x^{4} - 5x)}{h}\)
• A1 for a correct numerator, \(8x^{3}h + 12x^{2}h^{2} + 8xh^{3} + 2h^{4} - 5h\)
• A1 for dividing by \(h\) to obtain \(8x^{3} - 5 + 12x^{2}h + 8xh^{2} + 2h^{3}\)
• A1* for a complete proof: as \(h \to 0\), \(8x^{3} - 5 + 12x^{2}h + 8xh^{2} + 2h^{3} \to 8x^{3} - 5\), so the derivative is \(8x^{3} - 5\) (cso)

Worked solution: Let \(f(x) = 2x^{4} - 5x\).
Using \((x + h)^{4} = x^{4} + 4x^{3}h + 6x^{2}h^{2} + 4xh^{3} + h^{4}\):
\(\frac{f(x + h) - f(x)}{h} = \frac{2(x + h)^{4} - 5(x + h) - (2x^{4} - 5x)}{h} = \frac{8x^{3}h + 12x^{2}h^{2} + 8xh^{3} + 2h^{4} - 5h}{h} = 8x^{3} - 5 + 12x^{2}h + 8xh^{2} + 2h^{3}\)
So \(f'(x) = \lim_{h \to 0}\left(8x^{3} - 5 + 12x^{2}h + 8xh^{2} + 2h^{3}\right) = 8x^{3} - 5\).

• M1 for substituting \(x = 1\) into \(8x^{3} - 5\) to find the gradient, \(3\)
• M1 for finding \(y = -3\) and using \(y - y_{1} = m(x - x_{1})\) with \((1, -3)\) and their gradient
• A1 for \(y = 3x - 6\) (cao)

Worked solution: At \(x = 1\): \(y = -3\) and \(\frac{\mathrm{d}y}{\mathrm{d}x} = 3\).
\(y + 3 = 3(x - 1)\) so \(y = 3x - 6\).