6 Fig. 6 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 2 \arcsin x , - 1 \leqslant x \leqslant 1\).
Fig. 6 also shows the curve \(y = \mathrm { g } ( x )\), where \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\).
P is the point on the curve \(y = \mathrm { f } ( x )\) with \(x\)-coordinate \(\frac { 1 } { 2 }\).
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{7b77c646-2bc5-4166-b22e-3c1229abd722-3_711_693_466_685}
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\caption{Fig. 6}
\end{figure}
- Find the \(y\)-coordinate of P , giving your answer in terms of \(\pi\).
The point Q is the reflection of P in \(y = x\).
- Find \(\mathrm { g } ( x )\) and its derivative \(\mathrm { g } ^ { \prime } ( x )\). Hence determine the exact gradient of the curve \(y = \mathrm { g } ( x )\) at the point Q .
Write down the exact gradient of \(y = \mathrm { f } ( x )\) at the point P .