Derivative of inverse trig function

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 $$\left( 1 + x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 0$$

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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 $$\frac { \mathrm { d } } { \mathrm {~d} x } \left( \sec ^ { - 1 } x \right) = \frac { 1 } { \sqrt { x ^ { 4 } - x ^ { 2 } } }$$

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1. (a) Given that \(x \geqslant 1\), show that \(\sec ^ { - 1 } x = \cos ^ { - 1 } \left( \frac { 1 } { x } \right)\), and deduce that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \sec ^ { - 1 } x \right) = \frac { 1 } { x \sqrt { x ^ { 2 } - 1 } }\).
   (b) Use integration by parts to determine \(\int \sec ^ { - 1 } x \mathrm {~d} x\).
2. \includegraphics[max width=\textwidth, alt={}, center]{5d526fd9-72f8-42b1-b156-fd4a0c764c82-4_670_1029_1073_596} The diagram shows the curve \(S\) with equation \(y = \sec ^ { - 1 } x\) for \(x \geqslant 1\). The line \(L\), with gradient \(\frac { 1 } { \sqrt { 2 } }\), is the tangent to \(S\) at the point \(P\) and cuts the \(x\)-axis at the point \(Q\). The point \(I\) has coordinates \(( 1,0 )\).
   (a) Determine the exact coordinates of \(P\) and \(Q\).
   (b) The region \(R\), shaded on the diagram, is bounded by the line segments \(P Q\) and \(Q I\) and the \(\operatorname { arc } I P\) of \(S\). Show that \(R\) has area $$\ln ( 1 + \sqrt { 2 } ) - \frac { \pi ( 8 - \pi ) \sqrt { 2 } } { 32 } .$$ {www.cie.org.uk} after the live examination series. }

Given that \(y = \arctan \frac{x}{\sqrt{1 + x^2}}\) show that \(\frac{dy}{dx} = \frac{1}{\sqrt{1 + x^2}}\) [4]

The curve \(C\) has equation $$y = \operatorname{arcsec} e^x, \quad x > 0, \quad 0 < y < \frac{1}{2}\pi.$$

1. Prove that \(\frac{dy}{dx} = \frac{1}{\sqrt{e^{2x} - 1}}\). [5]
2. Sketch the graph of \(C\). [2]

The point \(A\) on \(C\) has \(x\)-coordinate \(\ln 2\). The tangent to \(C\) at \(A\) intersects the \(y\)-axis at the point \(B\).

1. Find the exact value of the \(y\)-coordinate of \(B\). [4]

Find the gradient of the tangent to the curve $$y = \sin^{-1} x$$ at the point where \(x = \frac{1}{5}\) Circle your answer. [1 mark] \(\frac{5\sqrt{6}}{12}\) \quad \(\frac{2\sqrt{6}}{5}\) \quad \(\frac{4\sqrt{3}}{25}\) \quad \(\frac{25}{24}\)