AQA C3 2006 June — Question 9

Exam BoardAQA
ModuleC3 (Core Mathematics 3)
Year2006
SessionJune
TopicDifferentiating Transcendental Functions

9 The diagram shows the curve with equation \(y = \sin ^ { - 1 } 2 x\), where \(- \frac { 1 } { 2 } \leqslant x \leqslant \frac { 1 } { 2 }\).
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  1. Find the \(y\)-coordinate of the point \(A\), where \(x = \frac { 1 } { 2 }\).
    1. Given that \(y = \sin ^ { - 1 } 2 x\), show that \(x = \frac { 1 } { 2 } \sin y\).
    2. Given that \(x = \frac { 1 } { 2 } \sin y\), find \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) in terms of \(y\).
  2. Using the answers to part (b) and a suitable trigonometrical identity, show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { \sqrt { 1 - 4 x ^ { 2 } } }$$