Home / A-Level Further Maths 9231 / Further differentiation / Differentiating hyperbolic and inverse functions
Cambridge International AS & A Level · Topic 2.3
A-Level Further Maths 9231 / Topic 2.3 / AS & A Level

Six hyperbolic derivatives, five inverse ones, one method.

The hyperbolic functions differentiate almost like the trigonometric ones, with a few signs changed. Their inverses, and \(\sin^{-1}\) and \(\cos^{-1}\), are differentiated by turning \(y = f^{-1}(x)\) round into \(f(y) = x\) and differentiating implicitly, with the range deciding the sign of a square root.

Topic 2.3All further differentiation lessons →
The key idea

\(\dfrac{d}{dx}\sinh x = \cosh x\), \(\dfrac{d}{dx}\cosh x = \sinh x\) (no minus sign), \(\dfrac{d}{dx}\tanh x = \operatorname{sech}^{2} x\). For an inverse, write \(y = f^{-1}(x)\) as \(f(y) = x\), differentiate implicitly and use the RANGE of \(y\) to choose the sign of the root: \(\dfrac{d}{dx}\sin^{-1} x = \dfrac{1}{\sqrt{1 - x^{2}}}\), \(\dfrac{d}{dx}\cos^{-1} x = -\dfrac{1}{\sqrt{1 - x^{2}}}\), \(\dfrac{d}{dx}\sinh^{-1} x = \dfrac{1}{\sqrt{x^{2} + 1}}\), \(\dfrac{d}{dx}\cosh^{-1} x = \dfrac{1}{\sqrt{x^{2} - 1}}\), \(\dfrac{d}{dx}\tanh^{-1} x = \dfrac{1}{1 - x^{2}}\).

x y O −1 1 2 4 8 12 ln 2 y = 5 cosh x − 3 sinh x minimum (ln 2, 4) dy/dx = 0 where tanh x = 3/5
Fig. 1: y = 5 cosh x - 3 sinh x. Its gradient 5 sinh x - 3 cosh x is zero only where tanh x = 3/5, that is x = tanh⁻¹(3/5) = ln 2, and the curve has its minimum value 4 there
Section 01

Hyperbolic derivatives and stationary points.

Differentiate the exponential definitions directly: \(\dfrac{d}{dx}\left(\dfrac{e^{x} - e^{-x}}{2}\right) = \dfrac{e^{x} + e^{-x}}{2}\), so \(\dfrac{d}{dx}\sinh x = \cosh x\), and in the same way \(\dfrac{d}{dx}\cosh x = \sinh x\). The other four follow from these two by the quotient or chain rule. Compared with trigonometry, the signs change for \(\cosh\) and \(\operatorname{sech}\): there is no minus in \(\dfrac{d}{dx}\cosh x\), but there IS one in \(\dfrac{d}{dx}\operatorname{sech} x\).

\(f(x)\)\(f'(x)\)\(f(x)\)\(f'(x)\)
\(\sinh x\)\(\cosh x\)\(\operatorname{cosech} x\)\(-\operatorname{cosech} x\coth x\)
\(\cosh x\)\(\sinh x\)\(\operatorname{sech} x\)\(-\operatorname{sech} x\tanh x\)
\(\tanh x\)\(\operatorname{sech}^{2} x\)\(\coth x\)\(-\operatorname{cosech}^{2} x\)

The chain, product and quotient rules work exactly as before: \(\dfrac{d}{dx}\cosh 3x = 3\sinh 3x\), \(\dfrac{d}{dx}\big(x^{2}\cosh 3x\big) = 2x\cosh 3x + 3x^{2}\sinh 3x\). For a stationary point of \(a\cosh x + b\sinh x\), set the derivative to zero and divide by \(\cosh x\) (never zero) to get an equation in \(\tanh x\).

Worked example Derive tanh and sech
Starting from \(\dfrac{d}{dx}\sinh x = \cosh x\) and \(\dfrac{d}{dx}\cosh x = \sinh x\), show that \(\dfrac{d}{dx}\tanh x = \operatorname{sech}^{2} x\) and \(\dfrac{d}{dx}\operatorname{sech} x = -\operatorname{sech} x\tanh x\).
  1. Quotient rule on \(\tanh x = \dfrac{\sinh x}{\cosh x}\):
  2. \[\begin{aligned}\frac{d}{dx}\tanh x &= \frac{\cosh x\cosh x - \sinh x\sinh x}{\cosh^{2} x}\\ &= \frac{\cosh^{2} x - \sinh^{2} x}{\cosh^{2} x} = \frac{1}{\cosh^{2} x}\end{aligned}\]
  3. using \(\cosh^{2} x - \sinh^{2} x = 1\). So \(\dfrac{d}{dx}\tanh x = \operatorname{sech}^{2} x\).
  4. Chain rule on \(\operatorname{sech} x = (\cosh x)^{-1}\): \(\dfrac{d}{dx}\operatorname{sech} x = -(\cosh x)^{-2}\sinh x = -\dfrac{1}{\cosh x}\cdot\dfrac{\sinh x}{\cosh x}\).
  5. That is \(-\operatorname{sech} x\tanh x\).
Answer: \(\dfrac{d}{dx}\tanh x = \operatorname{sech}^{2} x\) and \(\dfrac{d}{dx}\operatorname{sech} x = -\operatorname{sech} x\tanh x\)
Worked example A minimum via tanh x = 3/5
Find the exact coordinates of the stationary point of \(y = 5\cosh x - 3\sinh x\) and show that it is a minimum.
  1. \(\dfrac{dy}{dx} = 5\sinh x - 3\cosh x = 0\). Divide by \(\cosh x\), which is never zero: \(5\tanh x = 3\), so \(\tanh x = \tfrac{3}{5}\).
  2. \(|\tfrac{3}{5}| \lt 1\), so there is exactly one solution: \(x = \tanh^{-1}\tfrac{3}{5} = \tfrac{1}{2}\ln\!\left(\dfrac{1 + \frac{3}{5}}{1 - \frac{3}{5}}\right) = \tfrac{1}{2}\ln 4 = \ln 2\).
  3. \(\cosh(\ln 2) = \dfrac{2 + \frac{1}{2}}{2} = \dfrac{5}{4}\) and \(\sinh(\ln 2) = \dfrac{2 - \frac{1}{2}}{2} = \dfrac{3}{4}\).
  4. \(y = 5\cdot\tfrac{5}{4} - 3\cdot\tfrac{3}{4} = \tfrac{25}{4} - \tfrac{9}{4} = 4\).
  5. \(\dfrac{d^{2}y}{dx^{2}} = 5\cosh x - 3\sinh x = y\), which is \(4 \gt 0\) at this point, so it is a minimum (Fig. 1).
Answer: Minimum at \((\ln 2,\ 4)\); the minimum value is \(4\)
Your turn
Differentiate \(y = \ln(\cosh 2x)\).
Examiner trap

Do not copy the trigonometric signs. \(\dfrac{d}{dx}\cos x = -\sin x\), but \(\dfrac{d}{dx}\cosh x = +\sinh x\). And \(\dfrac{d}{dx}\sec x = +\sec x\tan x\), but \(\dfrac{d}{dx}\operatorname{sech} x = -\operatorname{sech} x\tanh x\). When in doubt, rebuild it from \(\cosh\) and \(\sinh\).

Section 02

Inverse trigonometric functions, implicitly.

To differentiate \(y = \sin^{-1} x\), turn it round: \(\sin y = x\). Differentiate both sides with respect to \(x\) and solve for \(\dfrac{dy}{dx}\). The answer comes out in terms of \(y\), and you convert it back to \(x\) with \(\cos^{2} y + \sin^{2} y = 1\). That identity gives a square root, and its SIGN is fixed by the range of the inverse function: \(\sin^{-1} x\) takes values in \(-\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2}\), \(\cos^{-1} x\) in \(0 \le y \le \pi\).

The results, valid for \(-1 \lt x \lt 1\) (at \(x = \pm 1\) the graphs have vertical tangents):

\[\frac{d}{dx}\sin^{-1} x = \frac{1}{\sqrt{1 - x^{2}}}, \qquad \frac{d}{dx}\cos^{-1} x = -\frac{1}{\sqrt{1 - x^{2}}}.\]

With the chain rule: \(\dfrac{d}{dx}\sin^{-1}(u) = \dfrac{u'}{\sqrt{1 - u^{2}}}\) and \(\dfrac{d}{dx}\cos^{-1}(u) = -\dfrac{u'}{\sqrt{1 - u^{2}}}\).

Worked example Derive both, with the sign
Show that \(\dfrac{d}{dx}\sin^{-1} x = \dfrac{1}{\sqrt{1 - x^{2}}}\) and \(\dfrac{d}{dx}\cos^{-1} x = -\dfrac{1}{\sqrt{1 - x^{2}}}\) for \(-1 \lt x \lt 1\).
  1. Let \(y = \sin^{-1} x\), so \(\sin y = x\). Differentiate implicitly: \(\cos y\,\dfrac{dy}{dx} = 1\), so \(\dfrac{dy}{dx} = \dfrac{1}{\cos y}\).
  2. \(\cos^{2} y = 1 - \sin^{2} y = 1 - x^{2}\). Since \(-\tfrac{\pi}{2} \lt y \lt \tfrac{\pi}{2}\), \(\cos y \gt 0\), so \(\cos y = +\sqrt{1 - x^{2}}\) and \(\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - x^{2}}}\).
  3. Let \(y = \cos^{-1} x\), so \(\cos y = x\). Then \(-\sin y\,\dfrac{dy}{dx} = 1\), so \(\dfrac{dy}{dx} = -\dfrac{1}{\sin y}\).
  4. Since \(0 \lt y \lt \pi\), \(\sin y \gt 0\), so \(\sin y = +\sqrt{1 - x^{2}}\) and \(\dfrac{dy}{dx} = -\dfrac{1}{\sqrt{1 - x^{2}}}\).
  5. Check: the two derivatives add to zero, which fits \(\sin^{-1} x + \cos^{-1} x = \tfrac{\pi}{2}\), a constant.
Answer: \(\dfrac{d}{dx}\sin^{-1} x = \dfrac{1}{\sqrt{1 - x^{2}}}\), \(\dfrac{d}{dx}\cos^{-1} x = -\dfrac{1}{\sqrt{1 - x^{2}}}\)
Worked example Chain and product rules
(a) Differentiate \(\sin^{-1}(3x)\). (b) Find the exact gradient of \(y = x\cos^{-1} x\) at \(x = \tfrac{1}{2}\).
  1. (a) With \(u = 3x\), \(u' = 3\): \(\dfrac{d}{dx}\sin^{-1}(3x) = \dfrac{3}{\sqrt{1 - (3x)^{2}}} = \dfrac{3}{\sqrt{1 - 9x^{2}}}\), for \(|x| \lt \tfrac{1}{3}\).
  2. (b) Product rule: \(\dfrac{dy}{dx} = \cos^{-1} x + x\cdot\left(-\dfrac{1}{\sqrt{1 - x^{2}}}\right) = \cos^{-1} x - \dfrac{x}{\sqrt{1 - x^{2}}}\).
  3. At \(x = \tfrac{1}{2}\): \(\cos^{-1}\tfrac{1}{2} = \tfrac{\pi}{3}\) and \(\sqrt{1 - \tfrac{1}{4}} = \tfrac{\sqrt{3}}{2}\), so the second term is \(\dfrac{1/2}{\sqrt{3}/2} = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}\).
  4. Gradient \(= \dfrac{\pi}{3} - \dfrac{\sqrt{3}}{3} = \dfrac{\pi - \sqrt{3}}{3}\) (about \(0.470\)).
Answer: (a) \(\dfrac{3}{\sqrt{1 - 9x^{2}}}\)   (b) \(\dfrac{\pi - \sqrt{3}}{3}\)
Your turn
Show that \(\dfrac{d}{dx}\sin^{-1}\!\left(\dfrac{x}{2}\right) = \dfrac{1}{\sqrt{4 - x^{2}}}\) for \(|x| \lt 2\).
Examiner trap

In a 'show that', the line '\(\cos y = \sqrt{1 - x^{2}}\)' needs its reason: \(\cos y \ge 0\) BECAUSE \(-\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2}\). For \(\cos^{-1}\) the minus sign comes from differentiating \(\cos y\), not from the root, which is positive because \(0 \le y \le \pi\).

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Differentiate \(\cosh 2x\).

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Which is \(\dfrac{d}{dx}\tanh x\)?

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Deriving \(\dfrac{d}{dx}\sin^{-1} x\), you reach \(\dfrac{dy}{dx} = \dfrac{1}{\cos y}\). Why is \(\cos y = +\sqrt{1 - x^{2}}\)?

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Which is \(\dfrac{d}{dx}\tanh^{-1} x\) for \(|x| \lt 1\)?

Section 03

Inverse hyperbolic functions, two ways.

The same method works for \(\sinh^{-1}\), \(\cosh^{-1}\) and \(\tanh^{-1}\), now using \(\cosh^{2} y - \sinh^{2} y = 1\) and \(\operatorname{sech}^{2} y = 1 - \tanh^{2} y\). Because section 2.1 gave each inverse a logarithmic form, you can also differentiate the log directly and get the same answer: a good check, and a question may ask for either method.

\(y\)Turned round\(\dfrac{dy}{dx}\)Valid for
\(\sinh^{-1} x\)\(\sinh y = x\)\(\dfrac{1}{\sqrt{x^{2} + 1}}\)all real \(x\)
\(\cosh^{-1} x\)\(\cosh y = x,\ y \ge 0\)\(\dfrac{1}{\sqrt{x^{2} - 1}}\)\(x \gt 1\)
\(\tanh^{-1} x\)\(\tanh y = x\)\(\dfrac{1}{1 - x^{2}}\)\(|x| \lt 1\)

For \(\cosh^{-1}\): \(\sinh y\,\dfrac{dy}{dx} = 1\) and \(\sinh^{2} y = x^{2} - 1\); the principal value \(y \ge 0\) makes \(\sinh y \ge 0\), so the root is positive. For \(\tanh^{-1}\): \(\operatorname{sech}^{2} y\,\dfrac{dy}{dx} = 1\), so \(\dfrac{dy}{dx} = \dfrac{1}{1 - \tanh^{2} y} = \dfrac{1}{1 - x^{2}}\), with no square root at all.

Worked example sinh⁻¹ x both ways
Show that \(\dfrac{d}{dx}\sinh^{-1} x = \dfrac{1}{\sqrt{x^{2} + 1}}\) (a) by implicit differentiation (b) from \(\sinh^{-1} x = \ln\!\big(x + \sqrt{x^{2} + 1}\big)\).
  1. (a) Let \(y = \sinh^{-1} x\), so \(\sinh y = x\) and \(\cosh y\,\dfrac{dy}{dx} = 1\).
  2. \(\cosh^{2} y = 1 + \sinh^{2} y = 1 + x^{2}\), and \(\cosh y \gt 0\) for every \(y\), so \(\cosh y = \sqrt{x^{2} + 1}\) and \(\dfrac{dy}{dx} = \dfrac{1}{\sqrt{x^{2} + 1}}\).
  3. (b) Chain rule on the log, with \(\dfrac{d}{dx}\sqrt{x^{2} + 1} = \dfrac{x}{\sqrt{x^{2} + 1}}\):
  4. \[\begin{aligned}\frac{dy}{dx} &= \frac{1 + \dfrac{x}{\sqrt{x^{2} + 1}}}{x + \sqrt{x^{2} + 1}}\\ &= \frac{\sqrt{x^{2} + 1} + x}{\sqrt{x^{2} + 1}\,\big(x + \sqrt{x^{2} + 1}\big)}\\ &= \frac{1}{\sqrt{x^{2} + 1}}\end{aligned}\]
  5. Both methods give the same result.
Answer: \(\dfrac{d}{dx}\sinh^{-1} x = \dfrac{1}{\sqrt{x^{2} + 1}}\)
Worked example Three chain rule versions
Differentiate (a) \(\sinh^{-1}(2x)\) (b) \(\cosh^{-1}(x^{2} + 1)\) for \(x \gt 0\) (c) \(\tanh^{-1}(\sin x)\) for \(-\tfrac{\pi}{2} \lt x \lt \tfrac{\pi}{2}\), simplifying each.
  1. (a) \(\dfrac{2}{\sqrt{(2x)^{2} + 1}} = \dfrac{2}{\sqrt{4x^{2} + 1}}\).
  2. (b) With \(u = x^{2} + 1\), \(u' = 2x\):
  3. \[\begin{aligned}\frac{dy}{dx} &= \frac{2x}{\sqrt{(x^{2} + 1)^{2} - 1}} = \frac{2x}{\sqrt{x^{4} + 2x^{2}}}\\ &= \frac{2x}{x\sqrt{x^{2} + 2}} = \frac{2}{\sqrt{x^{2} + 2}}\end{aligned}\]
  4. using \(\sqrt{x^{2}} = x\) because \(x \gt 0\) (for \(x \lt 0\) it would be \(-x\), giving \(-\dfrac{2}{\sqrt{x^{2} + 2}}\)).
  5. (c) With \(u = \sin x\), \(u' = \cos x\): \(\dfrac{\cos x}{1 - \sin^{2} x} = \dfrac{\cos x}{\cos^{2} x} = \dfrac{1}{\cos x} = \sec x\). (Here \(|\sin x| \lt 1\), so \(\tanh^{-1}(\sin x)\) is defined.)
Answer: (a) \(\dfrac{2}{\sqrt{4x^{2} + 1}}\)   (b) \(\dfrac{2}{\sqrt{x^{2} + 2}}\)   (c) \(\sec x\)
Your turn
Use \(\tanh^{-1} x = \tfrac{1}{2}\ln\!\left(\dfrac{1 + x}{1 - x}\right)\) to show that \(\dfrac{d}{dx}\tanh^{-1} x = \dfrac{1}{1 - x^{2}}\).
Examiner trap

Five similar-looking results: only \(\sinh^{-1}\) has \(x^{2} + 1\); \(\cosh^{-1}\) has \(x^{2} - 1\); \(\sin^{-1}\) and \(\cos^{-1}\) have \(1 - x^{2}\) under a root; \(\tanh^{-1}\) has \(1 - x^{2}\) with NO root. Writing \(\dfrac{1}{\sqrt{1 - x^{2}}}\) for \(\tanh^{-1}\) is the commonest slip. And with the chain rule, square the whole inner function: \((2x)^{2} = 4x^{2}\), not \(2x^{2}\).

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Differentiate \(\sinh^{-1}(3x)\) with respect to \(x\).

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Find the minimum value of \(5\cosh x + 4\sinh x\).

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Differentiate \(\cos^{-1}(2x)\) for \(|x| \lt \tfrac{1}{2}\).

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Find the exact gradient of \(y = x\sinh^{-1} x\) at the point where \(x = \tfrac{3}{4}\).

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Hyperbolic and inverse functions differentiated.

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Six original Cambridge-style questions
Three quick hyperbolic derivatives, deriving \(\coth\) and \(\operatorname{cosech}\), \(\cos^{-1}\) by implicit differentiation, two chain rule inverses that simplify, a maximum of \(4\sinh x - 5\cosh x\), and a second-order result for \(\sinh^{-1} x\).