Second derivative relations with hyperbolics

4 It is given that $$x = - t + \tan ^ { - 1 } t \quad \text { and } \quad y = t + \sinh ^ { - 1 } t$$

1. Show that \(\frac { d y } { d x } = - \frac { t ^ { 2 } + 1 + \sqrt { t ^ { 2 } + 1 } } { t ^ { 2 } }\).
2. Find the value of \(\frac { \mathrm { d } ^ { 2 } \mathrm { y } } { \mathrm { dx } ^ { 2 } }\) when \(t = \frac { 3 } { 4 }\).

5 It is given that $$x = \sinh ^ { - 1 } t , \quad y = \cos ^ { - 1 } t$$ where \(- 1 < t < 1\).

1. By differentiating \(\cos y\) with respect to \(t\), show that \(\frac { d y } { d t } = - \frac { 1 } { \sqrt { 1 - t ^ { 2 } } }\).
2. Find \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\) in terms of \(t\), simplifying your answer.

3. Given that $$y = x - \operatorname { artanh } \left( \frac { 2 x } { 1 + x ^ { 2 } } \right)$$

1. show that $$1 - \frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { k } { 1 - x ^ { 2 } }$$ where \(k\) is a constant to be found.
2. Hence, or otherwise, show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + x \left( 1 - \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } = 0$$

3. (a) Prove that $$\frac { \mathrm { d } ( \operatorname { arcoth } x ) } { \mathrm { d } x } = \frac { 1 } { 1 - x ^ { 2 } }$$ Given that \(y = ( \operatorname { arcoth } x ) ^ { 2 }\),
(b) show that $$\left( 1 - x ^ { 2 } \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } - 2 x \frac { d y } { d x } = \frac { k } { 1 - x ^ { 2 } }$$ where \(k\) is a constant to be determined.

6

1. Show that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \sinh ^ { - 1 } x \right) = \frac { 1 } { \sqrt { x ^ { 2 } + 1 } }\).
2. Given that \(y = \cosh \left( a \sinh ^ { - 1 } x \right)\), where \(a\) is a constant, show that $$\left( x ^ { 2 } + 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + x \frac { \mathrm {~d} y } { \mathrm {~d} x } - a ^ { 2 } y = 0$$

9 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = \ln ( 1 + \sinh x )\).

1. Given that \(k\) lies in the domain of this function, explain why \(k\) must be greater than \(\ln ( \sqrt { 2 } - 1 )\).
2. 1. Find \(\mathrm { f } ^ { \prime } ( x )\).
   2. Show that \(\mathrm { f } ^ { \prime \prime } ( \mathrm { x } ) = \frac { \mathrm { a } \sinh \mathrm { x } + \mathrm { b } } { ( 1 + \sinh \mathrm { x } ) ^ { 2 } }\), where \(a\) and \(b\) are integers to be determined.
3. Hence find a quadratic approximation to \(\mathrm { f } ( x )\) for small values of \(x\).
4. Find the percentage error in this approximation when \(x = 0.1\).

1. Given that

$$y = \mathrm { e } ^ { 2 x } \sinh x$$ prove by induction that for \(n \in \mathbb { N }\) $$\frac { \mathrm { d } ^ { n } y } { \mathrm {~d} x ^ { n } } = \mathrm { e } ^ { 2 x } \left( \frac { 3 ^ { n } + 1 } { 2 } \sinh x + \frac { 3 ^ { n } - 1 } { 2 } \cosh x \right)$$

5. Given that \(y = ( \operatorname { arcosh } 3 x ) ^ { 2 }\), where \(3 x > 1\), show that

1. \(\left( 9 x ^ { 2 } - 1 \right) \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } = 36 y\),
2. \(\left( 9 x ^ { 2 } - 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 9 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 18\).

3

1. Given that \(y = \sqrt { \sinh x }\) for \(x \geqslant 0\), express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(y\) only.
2. Find \(\int \frac { 2 t } { \sqrt { 1 + t ^ { 4 } } } \mathrm {~d} t\).

4

1. Given that \(y = \sqrt { \sinh x }\) for \(x \geqslant 0\), express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(y\) only.
2. Hence or otherwise find \(\int \frac { 2 t } { \sqrt { 1 + t ^ { 4 } } } \mathrm {~d} t\).

4

1. Given that \(y = \sqrt { \sinh x }\) for \(x \geqslant 0\), express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(y\) only.
2. Hence or otherwise find \(\int \frac { 2 t } { \sqrt { 1 + t ^ { 4 } } } \mathrm {~d} t\).

12 The curve \(C\) is defined parametrically by $$x = t + \ln ( \cosh t ) , \quad y = \sinh t$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { - t } \cosh ^ { 2 } t\).
2. Hence show that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \mathrm { e } ^ { - 2 t } \cosh ^ { 2 } t ( 2 \sinh t - \cosh t )\).
3. Find the exact value of \(t\) at the point on \(C\) where \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 0\).

Given that \(y = \sin x + \sinh x\), find \(\frac{d^2y}{dx^2} + y\) Circle your answer. [1 mark] \(2\sin x\) \quad \(-2\sin x\) \quad \(2\sinh x\) \quad \(-2\sinh x\)