A2 June 2020 Paper 2 Q3
3 Find the gradient of the tangent to the curve
\[y = \sin^{-1} x\]at the point where \(x = \dfrac{1}{5}\)
Circle your answer. [1 mark]
- \[\frac{5\sqrt{6}}{12}\]
- \[\frac{2\sqrt{6}}{5}\]
- \[\frac{4\sqrt{3}}{25}\]
- \[\frac{25}{24}\]
| Scheme | Marks | AO |
|---|---|---|
| Circles \(\dfrac{5\sqrt{6}}{12}\) | B1 | 1.1b |
| (1 mark) |