AS & A Level · Practice questions · Differentiating hyperbolic and inverse functions

Turn it round, then differentiate.

Six original Cambridge-style questions on differentiating hyperbolic functions and the inverse functions \(\sin^{-1}\), \(\cos^{-1}\), \(\sinh^{-1}\), \(\cosh^{-1}\) and \(\tanh^{-1}\), with stationary points and a proof.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Rebuild from \(\cosh\) and \(\sinh\); turn inverses round.

Hyperbolic: \(\sinh \to \cosh\), \(\cosh \to \sinh\) (no minus); the rest by quotient or chain rule.
Inverse: write \(y = f^{-1}(x)\) as \(f(y) = x\) and differentiate implicitly.
Back to \(x\): use \(\cos^{2} y + \sin^{2} y = 1\) or \(\cosh^{2} y - \sinh^{2} y = 1\), choosing the sign of the root from the RANGE of \(y\).
Chain rule: replace \(x\) by \(u\) and multiply by \(u'\); square the whole of \(u\).
Stationary points of \(a\cosh x + b\sinh x\): divide by \(\cosh x\) to get \(\tanh x = k\), which needs \(|k| \lt 1\).

Recall

[3 marks]

Differentiate with respect to \(x\): (a) \(\sinh 4x\) (b) \(\operatorname{sech} 2x\) (c) \(\cosh^{2} x\), giving (c) in terms of a single hyperbolic function.

(a) \(4\cosh 4x\). ✓
(b) \(-2\operatorname{sech} 2x\tanh 2x\). ✓
(c) \(2\cosh x\cdot\sinh x = \sinh 2x\), using \(\sinh 2x = 2\sinh x\cosh x\). ✓

Calculation

[4 marks]

Using only \(\dfrac{d}{dx}\sinh x = \cosh x\) and \(\dfrac{d}{dx}\cosh x = \sinh x\), show that (a) \(\dfrac{d}{dx}\coth x = -\operatorname{cosech}^{2} x\) (b) \(\dfrac{d}{dx}\operatorname{cosech} x = -\operatorname{cosech} x\coth x\).

(a) Quotient rule on \(\coth x = \dfrac{\cosh x}{\sinh x}\): ✓
\[\begin{aligned}\frac{d}{dx}\coth x &= \frac{\sinh x\sinh x - \cosh x\cosh x}{\sinh^{2} x}\\ &= -\frac{\cosh^{2} x - \sinh^{2} x}{\sinh^{2} x} = -\frac{1}{\sinh^{2} x}\end{aligned}\] ✓
So \(\dfrac{d}{dx}\coth x = -\operatorname{cosech}^{2} x\). ✓
(b) \(\operatorname{cosech} x = (\sinh x)^{-1}\), so its derivative is \(-(\sinh x)^{-2}\cosh x = -\dfrac{1}{\sinh x}\cdot\dfrac{\cosh x}{\sinh x} = -\operatorname{cosech} x\coth x\). ✓

Calculation

[4 marks]

(a) By writing \(y = \cos^{-1} x\) as \(\cos y = x\), show that \(\dfrac{dy}{dx} = -\dfrac{1}{\sqrt{1 - x^{2}}}\) for \(-1 \lt x \lt 1\), explaining the sign of the square root.

(b) Hence differentiate \(\cos^{-1}(x^{2})\) for \(0 \lt x \lt 1\).

(a) \(-\sin y\,\dfrac{dy}{dx} = 1\), so \(\dfrac{dy}{dx} = -\dfrac{1}{\sin y}\). ✓
\(\sin^{2} y = 1 - \cos^{2} y = 1 - x^{2}\). The range of \(\cos^{-1}\) is \(0 \le y \le \pi\), where \(\sin y \ge 0\) (and \(\sin y \gt 0\) for \(-1 \lt x \lt 1\)), so \(\sin y = +\sqrt{1 - x^{2}}\). ✓
Hence \(\dfrac{dy}{dx} = -\dfrac{1}{\sqrt{1 - x^{2}}}\). ✓
(b) With \(u = x^{2}\), \(u' = 2x\): \(\dfrac{d}{dx}\cos^{-1}(x^{2}) = -\dfrac{2x}{\sqrt{1 - (x^{2})^{2}}} = -\dfrac{2x}{\sqrt{1 - x^{4}}}\). ✓

Analysis

[5 marks]

(a) Show that \(\dfrac{d}{dx}\sinh^{-1}(\sqrt{x}) = \dfrac{1}{2\sqrt{x + x^{2}}}\) for \(x \gt 0\).

(b) Show that \(\dfrac{d}{dx}\tanh^{-1}(\cos x) = -\operatorname{cosec} x\) for \(0 \lt x \lt \pi\).

(a) With \(u = \sqrt{x}\), \(u' = \dfrac{1}{2\sqrt{x}}\) and \(u^{2} + 1 = x + 1\): ✓
\[\frac{d}{dx}\sinh^{-1}(\sqrt{x}) = \frac{1}{2\sqrt{x}\,\sqrt{x + 1}} = \frac{1}{2\sqrt{x(x + 1)}} = \frac{1}{2\sqrt{x + x^{2}}}\] ✓
(b) For \(0 \lt x \lt \pi\), \(|\cos x| \lt 1\), so \(\tanh^{-1}(\cos x)\) is defined. With \(u = \cos x\), \(u' = -\sin x\): ✓
\[\frac{d}{dx}\tanh^{-1}(\cos x) = \frac{-\sin x}{1 - \cos^{2} x} = \frac{-\sin x}{\sin^{2} x} = -\frac{1}{\sin x}\] ✓
which is \(-\operatorname{cosec} x\) (dividing by \(\sin x\) is fine, as \(\sin x \gt 0\) on \(0 \lt x \lt \pi\)). ✓

Analysis

[6 marks]

The curve \(C\) has equation \(y = 4\sinh x - 5\cosh x\).

(a) Show that \(C\) has exactly one stationary point, and find its exact coordinates.

(b) Determine its nature, and state the greatest value of \(4\sinh x - 5\cosh x\).

(a) \(\dfrac{dy}{dx} = 4\cosh x - 5\sinh x = 0\). Divide by \(\cosh x \neq 0\): \(\tanh x = \tfrac{4}{5}\). ✓
\(\tanh\) is one-one with \(|\tfrac{4}{5}| \lt 1\), so there is exactly one solution: \(x = \tanh^{-1}\tfrac{4}{5} = \tfrac{1}{2}\ln\!\left(\dfrac{9/5}{1/5}\right) = \tfrac{1}{2}\ln 9 = \ln 3\). ✓
\(\sinh(\ln 3) = \dfrac{3 - \frac{1}{3}}{2} = \dfrac{4}{3}\) and \(\cosh(\ln 3) = \dfrac{3 + \frac{1}{3}}{2} = \dfrac{5}{3}\). ✓
\(y = 4\cdot\tfrac{4}{3} - 5\cdot\tfrac{5}{3} = \tfrac{16 - 25}{3} = -3\). The stationary point is \((\ln 3,\ -3)\). ✓
(b) \(\dfrac{d^{2}y}{dx^{2}} = 4\sinh x - 5\cosh x = y = -3 \lt 0\), so it is a maximum. ✓
Being the only stationary point, it gives the greatest value: \(4\sinh x - 5\cosh x \le -3\) for all \(x\). ✓

Proof

[6 marks]

(a) Starting from \(\cosh^{-1} x = \ln\!\big(x + \sqrt{x^{2} - 1}\big)\), prove that \(\dfrac{d}{dx}\cosh^{-1} x = \dfrac{1}{\sqrt{x^{2} - 1}}\) for \(x \gt 1\).

(b) Given that \(y = \sinh^{-1} x\), prove that \((1 + x^{2})\dfrac{d^{2}y}{dx^{2}} + x\dfrac{dy}{dx} = 0\).

(a) Chain rule, with \(\dfrac{d}{dx}\sqrt{x^{2} - 1} = \dfrac{x}{\sqrt{x^{2} - 1}}\): ✓
\[\begin{aligned}\frac{d}{dx}\cosh^{-1} x &= \frac{1 + \dfrac{x}{\sqrt{x^{2} - 1}}}{x + \sqrt{x^{2} - 1}}\\ &= \frac{\sqrt{x^{2} - 1} + x}{\sqrt{x^{2} - 1}\,\big(x + \sqrt{x^{2} - 1}\big)}\\ &= \frac{1}{\sqrt{x^{2} - 1}}\end{aligned}\] ✓
(b) \(\dfrac{dy}{dx} = (1 + x^{2})^{-1/2}\). ✓
\(\dfrac{d^{2}y}{dx^{2}} = -\tfrac{1}{2}(1 + x^{2})^{-3/2}\cdot 2x = -x(1 + x^{2})^{-3/2}\). ✓
So \((1 + x^{2})\dfrac{d^{2}y}{dx^{2}} = -x(1 + x^{2})^{-1/2} = -x\dfrac{dy}{dx}\). ✓
Hence \((1 + x^{2})\dfrac{d^{2}y}{dx^{2}} + x\dfrac{dy}{dx} = 0\), as required. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.