A curve is defined parametrically by y = \sqrt{1 - t} and x = \cos^{-1} t, where 0 \le t < 1.
Show that \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\sqrt{1 + t}}{2}.
The mark scheme is shown once you've answered.
Practise this question\frac{\mathrm{d}y}{\mathrm{d}x} if y = \sin^2 5x + \sin^2 3x + \cos^2 3x.[3 marks]\frac{\mathrm{d}y}{\mathrm{d}x} if y = \sqrt{\cos x^2}.[4 marks]\frac{\mathrm{d}y}{\mathrm{d}x} if y = x^x.[4 marks]y = \cos^{-1} x, where 0 \le \cos^{-1} x \le \pi, prove that \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{\sqrt{1 - x^2}}.[7 marks]\frac{\mathrm{d}^2y}{\mathrm{d}x^2} in terms of t, giving your answer in simplified form.[3 marks]More practice: the rest of this paper · more Differentiation II questions · all CAPE Pure Mathematics Unit 2 past papers