Prove that the derivative of \(\cos ^ { - 1 } x\) is \(- \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }\).
A curve has equation \(y = \cos ^ { - 1 } \left( 1 - x ^ { 2 } \right)\), for \(0 < x < \sqrt { 2 }\).
Find and simplify \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), and hence show that
$$\left( 2 - x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = x \frac { \mathrm {~d} y } { \mathrm {~d} x }$$
Given that \(y = \sinh ^ { - 1 } x\), prove that \(y = \ln \left( x + \sqrt { x ^ { 2 } + 1 } \right)\).
It is given that \(x\) satisfies the equation \(\sinh ^ { - 1 } x - \cosh ^ { - 1 } x = \ln 2\). Use the logarithmic forms for \(\sinh ^ { - 1 } x\) and \(\cosh ^ { - 1 } x\) to show that
$$\sqrt { x ^ { 2 } + 1 } - 2 \sqrt { x ^ { 2 } - 1 } = x$$
Hence, by squaring this equation, find the exact value of \(x\).