Derivative of inverse trig function

5. Given that \(y = \arcsin x\) prove that

1. \(\quad \frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { \left( 1 - x ^ { 2 } \right) } }\)
2. \(\quad \left( 1 - x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 0\)

2

1. Given that \(y = \tan ^ { - 1 } x\), prove that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 1 + x ^ { 2 } }\).
2. Verify that \(y = \tan ^ { - 1 } x\) satisfies the equation $$\left( 1 + x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 0$$

1

1. 1. Differentiate the equation \(\sin y = x\) with respect to \(x\), and hence show that the derivative of \(\arcsin x\) is \(\frac { 1 } { \sqrt { 1 - x ^ { 2 } } }\).
   2. Evaluate the following integrals, giving your answers in exact form.
      (A) \(\int _ { - 1 } ^ { 1 } \frac { 1 } { \sqrt { 2 - x ^ { 2 } } } \mathrm {~d} x\)
      (B) \(\int _ { - \frac { 1 } { 2 } } ^ { \frac { 1 } { 2 } } \frac { 1 } { \sqrt { 1 - 2 x ^ { 2 } } } \mathrm {~d} x\)
2. A curve has polar equation \(r = \tan \theta , 0 \leqslant \theta < \frac { 1 } { 2 } \pi\). The points on the curve have cartesian coordinates \(( x , y )\). A sketch of the curve is given in Fig. 1. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{99f0c663-bb5b-4456-854c-df177f5d8349-2_493_796_1123_605} \captionsetup{labelformat=empty} \caption{Fig. 1}
   \end{figure} Show that \(x = \sin \theta\) and that \(r ^ { 2 } = \frac { x ^ { 2 } } { 1 - x ^ { 2 } }\).
   Hence show that the cartesian equation of the curve is $$y = \frac { x ^ { 2 } } { \sqrt { 1 - x ^ { 2 } } } .$$ Give the cartesian equation of the asymptote of the curve.

1 It is given that \(\mathrm { f } ( x ) = \tan ^ { - 1 } 2 x\) and \(\mathrm { g } ( x ) = p \tan ^ { - 1 } x\), where \(p\) is a constant. Find the value of \(p\) for which \(\mathrm { f } ^ { \prime } \left( \frac { 1 } { 2 } \right) = \mathrm { g } ^ { \prime } \left( \frac { 1 } { 2 } \right)\).

5 The function f , where \(\mathrm { f } ( x ) = \sec x\), has domain \(0 \leqslant x < \frac { \pi } { 2 }\) and has inverse function \(\mathrm { f } ^ { - 1 }\), where \(\mathrm { f } ^ { - 1 } ( x ) = \sec ^ { - 1 } x\).

1. Show that $$\sec ^ { - 1 } x = \cos ^ { - 1 } \frac { 1 } { x }$$
2. Hence show that $$\frac { \mathrm { d } } { \mathrm {~d} x } \left( \sec ^ { - 1 } x \right) = \frac { 1 } { \sqrt { x ^ { 4 } - x ^ { 2 } } }$$