P7.4aProduct, quotient and chain rules

(a) Answer: \(\frac{\mathrm{d}y}{\mathrm{d}x} = -21(-3x + 2)^{6}\)

• M1 for a derivative of the form \(k(-3x + 2)^{6}\)
• A1 for \(\frac{\mathrm{d}y}{\mathrm{d}x} = -21(-3x + 2)^{6}\) oe

Worked solution: \(\frac{\mathrm{d}y}{\mathrm{d}x} = 7(-3x + 2)^{6} \times (-3) = -21(-3x + 2)^{6}\)

(b) Answer: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4x}{2x^{2} + 1}\)

• M1 for a derivative of the form \(\frac{kx}{2x^{2} + 1}\)
• A1 for \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4x}{2x^{2} + 1}\) oe

Worked solution: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2x^{2} + 1} \times 4x = \frac{4x}{2x^{2} + 1}\)

(a) Answer: \(\frac{\mathrm{d}y}{\mathrm{d}x} = (2x - 4)\mathrm{e}^{x^{2} - 4x}\)

• M1 for a derivative of the form \((\alpha x + \beta)\mathrm{e}^{x^{2} - 4x}\)
• A1 for \(\frac{\mathrm{d}y}{\mathrm{d}x} = (2x - 4)\mathrm{e}^{x^{2} - 4x}\)

Worked solution: By the chain rule, \(\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{x^{2} - 4x} \times \frac{\mathrm{d}}{\mathrm{d}x}\left(x^{2} - 4x\right) = (2x - 4)\mathrm{e}^{x^{2} - 4x}\).

(b) Answer: \(\left(2, \mathrm{e}^{-4}\right)\)

• M1 for setting their \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and using \(\mathrm{e}^{x^{2} - 4x} \gt 0\) (or \(\ne 0\))
• A1 for \(x = 2\)
• A1 for \(\left(2, \mathrm{e}^{-4}\right)\)

Worked solution: \(\mathrm{e}^{x^{2} - 4x} \gt 0\) for all \(x\), so \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) only when \(2x - 4 = 0\), i.e. \(x = 2\).
Then \(y = \mathrm{e}^{4 - 8} = \mathrm{e}^{-4}\).

(c) Answer: Minimum

• M1 for \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \left(2 + (2x - 4)^{2}\right)\mathrm{e}^{x^{2} - 4x}\) considered at their \(x\), or a sign test on \(\frac{\mathrm{d}y}{\mathrm{d}x}\)
• A1 for \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 2\mathrm{e}^{-4} \gt 0\), so a minimum

Worked solution: \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 2\mathrm{e}^{x^{2} - 4x} + (2x - 4)^{2}\mathrm{e}^{x^{2} - 4x}\). At \(x = 2\) this is \(2\mathrm{e}^{-4} \gt 0\), so the stationary point is a minimum.

(a) Answer: \(y' = \mathrm{e}^x(\sin x + \cos x)\), \(y'' = 2\mathrm{e}^x\cos x\); \(2\mathrm{e}^x\cos x - 2\mathrm{e}^x(\sin x + \cos x) + 2\mathrm{e}^x\sin x = 0\)

• M1 for using the product rule: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^x\sin x + \mathrm{e}^x\cos x\)
• M1 for differentiating again with the product rule
• A1 for \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2\mathrm{e}^x\cos x\)
• A1* for substituting and simplifying to 0

Worked solution: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^x\sin x + \mathrm{e}^x\cos x = \mathrm{e}^x(\sin x + \cos x)\).
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \mathrm{e}^x(\sin x + \cos x) + \mathrm{e}^x(\cos x - \sin x) = 2\mathrm{e}^x\cos x\).
So \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2y = 2\mathrm{e}^x\cos x - 2\mathrm{e}^x\sin x - 2\mathrm{e}^x\cos x + 2\mathrm{e}^x\sin x = 0\).

(b) Answer: \(x = \frac{3\pi}{4}\) and \(x = \frac{7\pi}{4}\)

• M1 for \(\mathrm{e}^x(\sin x + \cos x) = 0 \Rightarrow \tan x = -1\) (with \(\mathrm{e}^x \ne 0\))
• A1 for \(\frac{3\pi}{4}\) and \(\frac{7\pi}{4}\) only

Worked solution: \(\mathrm{e}^x \gt 0\), so \(\sin x + \cos x = 0\), i.e. \(\tan x = -1\): \(x = \frac{3\pi}{4}\) or \(x = \frac{7\pi}{4}\).

(c) Answer: At a stationary point \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -2y\). At \(x = \frac{3\pi}{4}\), \(y \gt 0\), so a maximum; at \(x = \frac{7\pi}{4}\), \(y \lt 0\), so a minimum.

• M1 for \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -2y\)
• M1 for finding the sign of \(y\) at each point: \(y = \mathrm{e}^{\frac{3\pi}{4}} \times \frac{\sqrt{2}}{2} \gt 0\) and \(y = -\mathrm{e}^{\frac{7\pi}{4}} \times \frac{\sqrt{2}}{2} \lt 0\)
• A1 for maximum at \(x = \frac{3\pi}{4}\), minimum at \(x = \frac{7\pi}{4}\)

Worked solution: At a stationary point \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\), so part (a) gives \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -2y\).
At \(x = \frac{3\pi}{4}\): \(\sin x = \frac{\sqrt{2}}{2} \gt 0\), so \(y \gt 0\), \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \lt 0\): a maximum.
At \(x = \frac{7\pi}{4}\): \(\sin x = -\frac{\sqrt{2}}{2}\), so \(y \lt 0\), \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \gt 0\): a minimum.