Why is "Writing $\frac{d}{dx}\cosh x = -\sinh x$ (inventing a minus sign)" a common mistake in Differentiation of Hyperbolic and Inverse Trigonometric Functions?

Why it happens: Carrying over the circular result d/dx x = - x to the hyperbolic case by analogy. How to avoid it: Memorise: BOTH d/dx x = x and d/dx x = x are positive. The sign change only happens for circular functions. Source: 9231 Paper 2 Examiner Reports — recurring comment

Why is "Confusing $\sqrt{x^2+1}$ ($\sinh^{-1}$) with $\sqrt{x^2-1}$ ($\cosh^{-1}$) under the root" a common mistake in Differentiation of Hyperbolic and Inverse Trigonometric Functions?

Why it happens: The two inverse-hyperbolic derivatives look almost identical; the 1 is easy to misremember. How to avoid it: Link it to domains: ^-1x needs x ≥ 1, so √(x^2-1) (real only there); ^-1x is defined for all x, so √(x^2+1) (always real). Or just read it off MF19. Source: 9231 Paper 2 mark schemes

Why is "Omitting the $\frac{du}{dx}$ chain-rule factor, e.g. $\frac{d}{dx}\tan^{-1}(5x) = \frac{1}{1+25x^2}$" a common mistake in Differentiation of Hyperbolic and Inverse Trigonometric Functions?

Why it happens: Quoting the bare standard derivative and forgetting the inner function must be differentiated too. How to avoid it: Always write the standard derivative with u, then multiply by du/dx. For ^-1(5x) the answer is 5/1+25x^2. Source: 9231 Paper 2 Examiner Reports

Why is "Forming $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$ w.r.t. $t$ WITHOUT dividing by $\frac{dx}{dt}$" a common mistake in Differentiation of Hyperbolic and Inverse Trigonometric Functions?

Why it happens: Treating the parameter t as if it were x; forgetting the extra chain-rule step. How to avoid it: Use d^2y/dx^2 = d/dt\!(dy/dx)÷dx/dt. Differentiate, THEN divide by dx/dt once. Source: 9231 Paper 2 — parametric questions

Why is "Dropping the factorial denominators when assembling the Maclaurin series" a common mistake in Differentiation of Hyperbolic and Inverse Trigonometric Functions?

Why it happens: Writing f(0) + f'(0)x + f''(0)x^2 + ·s without the 1/2!, 1/3!, factors. How to avoid it: The coefficient of x^n is f^(n)(0)n!. Write the full template with factorials before substituting derivative values. Source: 9231 Paper 2 Examiner Reports

Why is "Stopping at the first 'three terms' when some coefficients are zero (e.g. for an odd function)" a common mistake in Differentiation of Hyperbolic and Inverse Trigonometric Functions?

Why it happens: A request for three NON-ZERO terms is read as three terms, so the student stops after vanishing coefficients. How to avoid it: Read the demand carefully: keep differentiating until you have the required number of NON-ZERO terms. Odd/even functions have alternating zero coefficients. Source: 9231 Paper 2 — series questions

Cambridge International A Levels Further Mathematics (9231)

Differentiation of Hyperbolic and Inverse Trigonometric Functions

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Detailed Study Notes

Detailed notes on Differentiation for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Differentiation of Hyperbolic & Inverse Trigonometric Functions — Cambridge International AS & A Level Further Mathematics 9231 Paper 2 (2026-2027 syllabus)

Differentiate hyperbolic, inverse trigonometric and inverse hyperbolic functions, find second derivatives from implicit and parametric relations, and build a Maclaurin series term-by-term using successive differentiation.

At a glance

Hyperbolic derivatives mirror the circular ones — but $\frac{d}{dx}\cosh x = \sinh x$ has NO minus sign.
Inverse trig & inverse hyperbolic derivatives ( $\sin^{-1}$ , $\tan^{-1}$ , $\sinh^{-1}$ , $\cosh^{-1}$ , $\tanh^{-1}$ ) ARE in the MF19 table — combine them with the chain, product and quotient rules.
Second derivatives come from implicit relations (differentiate twice, substitute $\frac{dy}{dx}$ back) or parametric ones ( $\frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy}{dx})\div\frac{dx}{dt}$ ).
Maclaurin series must be DERIVED: compute $f(0), f'(0), f''(0), \ldots$ by successive differentiation, then build $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots$ .
State the range of validity of any series you derive.

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

2.3a — Differentiate hyperbolic functions and the inverse trigonometric and inverse hyperbolic functions, using the standard derivatives in MF19 together with the chain, product and quotient rules.
2.3b — Find and use the second derivative of a function defined implicitly or parametrically.
2.3c — Derive the first few terms of the Maclaurin series of a function by successive differentiation and state the range of validity where appropriate.

Differentiating hyperbolic functions

sinh ↔ cosh swap cleanly — and unlike circular functions, no minus sign appears.

The three core hyperbolic functions differentiate in an almost circular-looking pattern, but with one crucial difference: no sign change between $\sinh$ and $\cosh$ .

The core three: $\frac{d}{dx}\sinh x = \cosh x, \qquad \frac{d}{dx}\cosh x = \sinh x, \qquad \frac{d}{dx}\tanh x = \text{sech}^2 x$

The reciprocal three (derive with the quotient rule, or quote from MF19):

$\frac{d}{dx}\text{sech}\,x = -\text{sech}\,x\tanh x$
$\frac{d}{dx}\text{cosech}\,x = -\text{cosech}\,x\coth x$
$\frac{d}{dx}\coth x = -\text{cosech}^2 x$

Why no minus on cosh? Because $\cosh x = \frac{e^x + e^{-x}}{2}$ , so $\frac{d}{dx}\cosh x = \frac{e^x - e^{-x}}{2} = \sinh x$ — both stay positive. Contrast circular: $\frac{d}{dx}\cos x = -\sin x$ .

The three standard hyperbolic derivatives. All are positive — the defining contrast with circular functions.

Cambridge tip. Whenever the inner function is more than just $x$ (e.g. $\cosh(3x^2)$ ), you MUST attach the chain-rule factor $\frac{du}{dx}$ . The bare standard derivative alone is an automatic mark loss.

$\frac{d}{dx}\sinh x = \cosh x$ and $\frac{d}{dx}\cosh x = \sinh x$ — both positive.
$\frac{d}{dx}\tanh x = \text{sech}^2 x$ .
Always attach the chain-rule factor for a non-trivial inner function.

See the full worked example for differentiation of hyperbolic and inverse trigonometric functions →

Inverse trigonometric & inverse hyperbolic derivatives

These standard results are in MF19 — your job is to apply them with the chain, product and quotient rules.

The inverse functions differentiate to neat algebraic fractions, and crucially all of these appear in the MF19 differentiation table — you do not need to memorise them, but you must apply them accurately.

Inverse trigonometric: $\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$

Inverse hyperbolic (watch the sign under the square root): $\frac{d}{dx}\sinh^{-1}x = \frac{1}{\sqrt{x^2+1}}, \quad \frac{d}{dx}\cosh^{-1}x = \frac{1}{\sqrt{x^2-1}}, \quad \frac{d}{dx}\tanh^{-1}x = \frac{1}{1-x^2}$

The sign pattern that catches people out:

$\sin^{-1}$ : $\sqrt{1-x^2}$ — defined for $|x| < 1$ .
$\sinh^{-1}$ : $\sqrt{x^2+1}$ — defined for all $x$ (always real).
$\cosh^{-1}$ : $\sqrt{x^2-1}$ — defined for $x \geq 1$ .

A handy memory link: the domain tells you the sign. $\cosh^{-1}$ only makes sense for $x \geq 1$ , so $\sqrt{x^2 - 1}$ is real exactly there.

Combining with the rules of differentiation. Exam questions almost always wrap these in a chain, product or quotient:

Chain: $\frac{d}{dx}\tan^{-1}(5x) = \frac{1}{1+(5x)^2}\cdot 5 = \frac{5}{1+25x^2}$ .
Product: $\frac{d}{dx}\big[x^2\cosh^{-1}x\big] = 2x\cosh^{-1}x + \frac{x^2}{\sqrt{x^2-1}}$ .
Quotient: for $\frac{\sin^{-1}x}{x}$ , apply $\frac{u'v - uv'}{v^2}$ with $u' = \frac{1}{\sqrt{1-x^2}}$ .

All five inverse derivatives sit in the MF19 differentiation table. The exam skill is applying them through the chain, product and quotient rules.

Cambridge tip. Never try to "simplify" $\sin^{-1}x$ or $\cosh^{-1}x$ away — they are functions in their own right. Quote the MF19 derivative exactly, attach $\frac{du}{dx}$ , then assemble.

$\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}$ , $\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$ .
$\sinh^{-1}$ has $\sqrt{x^2+1}$ ; $\cosh^{-1}$ has $\sqrt{x^2-1}$ .
All five are in MF19 — apply with chain/product/quotient rules.

See the full worked example for differentiation of hyperbolic and inverse trigonometric functions →

Second derivatives from implicit & parametric relations

Differentiate twice for implicit; differentiate dy/dx w.r.t. t then divide by dx/dt for parametric.

Higher-mark questions often ask for $\frac{d^2y}{dx^2}$ where $y$ is not given explicitly as a function of $x$ .

Implicit relations. Differentiate the equation twice with respect to $x$ , remembering that $y$ is a function of $x$ (so $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ ). After the second differentiation, substitute $\frac{dy}{dx}$ back in to express the answer cleanly.

Example sketch. For $x^2 + y^2 = 25$ : first $\frac{dy}{dx} = -\frac{x}{y}$ ; differentiating again and substituting gives $\frac{d^2y}{dx^2} = -\frac{25}{y^3}$ .

Parametric relations. When $x = x(t)$ and $y = y(t)$ :

Step 1 — first derivative: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

Step 2 — second derivative (the step everyone gets wrong): $\frac{d^2y}{dx^2} = \frac{d}{dt}\!\left(\frac{dy}{dx}\right)\div\frac{dx}{dt}$

You differentiate the first derivative with respect to $t$ , then divide by $\frac{dx}{dt}$ once — because of the chain rule $\frac{d}{dx} = \frac{1}{dx/dt}\cdot\frac{d}{dt}$ .

Worked logic. For $x = \sinh t$ , $y = \cosh 2t$ : $\frac{dy}{dx} = \frac{2\sinh 2t}{\cosh t} = 4\sinh t$ (using $\sinh 2t = 2\sinh t\cosh t$ ). Then $\frac{d^2y}{dx^2} = \frac{4\cosh t}{\cosh t} = 4$ .

Cambridge tip. It is almost always worth simplifying $\frac{dy}{dx}$ with a hyperbolic or trig identity BEFORE differentiating again — clean algebra prevents sign and quotient-rule errors at the second stage.

Implicit: differentiate twice, then substitute $\frac{dy}{dx}$ back in.
Parametric: $\frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy}{dx})\div\frac{dx}{dt}$ — divide by $\frac{dx}{dt}$ once.
Simplify $\frac{dy}{dx}$ before the second differentiation.

See the full worked example for differentiation of hyperbolic and inverse trigonometric functions →

Maclaurin series by successive differentiation

Compute f(0), f'(0), f''(0), … then assemble the series with factorial denominators.

A Maclaurin series expands $f(x)$ as a power series about $x = 0$ : $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$

The coefficient of $x^n$ is $\dfrac{f^{(n)}(0)}{n!}$ . You must derive these by successive differentiation — only a handful of standard series are in MF19, and the exam routinely asks for functions that are not.

The four-step method:

Differentiate successively to get $f'(x), f''(x), f'''(x), \ldots$ (simplify using hyperbolic identities like $\text{sech}^2 x = 1 - \tanh^2 x$ to keep the algebra manageable).
Evaluate each at $x = 0$ , using $\sinh 0 = 0$ , $\cosh 0 = 1$ , $\tanh 0 = 0$ , $\text{sech}\,0 = 1$ .
Substitute into the Maclaurin template, keeping the factorial denominators.
State the range of validity.

Worked logic — $\tanh x$ . $f'(x) = \text{sech}^2 x$ , $f''(x) = -2\tanh x\,\text{sech}^2 x$ , $f'''(x) = -2\text{sech}^2 x + 6\tanh^2 x\,\text{sech}^2 x$ . At $0$ : $f(0) = 0$ , $f'(0) = 1$ , $f''(0) = 0$ , $f'''(0) = -2$ . So $\tanh x = x - \frac{1}{3}x^3 + \cdots \quad (|x| < \tfrac{\pi}{2}).$

Notice the even-power coefficients vanish ( $\tanh$ is odd) — so reaching three NON-ZERO terms means differentiating further than you might expect.

The cubic Maclaurin polynomial (red dashed) hugs y = tanh x near the origin and drifts away further out — a picture of the "range of validity".

Cambridge tip. Always write the template with factorials first, then drop in your $f^{(n)}(0)$ values. The most-penalised error is omitting the $\frac{1}{2!}, \frac{1}{3!}, \ldots$ denominators.

Coefficient of $x^n$ is $\frac{f^{(n)}(0)}{n!}$ — keep the factorials.
Derive the coefficients by successive differentiation; don't just quote.
For odd/even functions, alternate coefficients vanish — go further for NON-ZERO terms.

See the full worked example for differentiation of hyperbolic and inverse trigonometric functions →

Quick recap

Hyperbolic: $\frac{d}{dx}\sinh x = \cosh x$ , $\frac{d}{dx}\cosh x = \sinh x$ (no minus), $\frac{d}{dx}\tanh x = \text{sech}^2 x$ .
Inverse derivatives are in MF19: $\sin^{-1}$ → $\frac{1}{\sqrt{1-x^2}}$ ; $\sinh^{-1}$ → $\frac{1}{\sqrt{x^2+1}}$ ; $\cosh^{-1}$ → $\frac{1}{\sqrt{x^2-1}}$ .
Implicit second derivative: differentiate twice, substitute $\frac{dy}{dx}$ back in.
Parametric second derivative: $\frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy}{dx})\div\frac{dx}{dt}$ .
Maclaurin: derive $f^{(n)}(0)$ by successive differentiation; coefficient of $x^n$ is $\frac{f^{(n)}(0)}{n!}$ .
Always attach the chain-rule factor, and state the range of validity of a series.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Hyperbolic derivatives $\frac{d}{dx}\sinh x = \cosh x$ , $\frac{d}{dx}\cosh x = \sinh x$ , $\frac{d}{dx}\tanh x = \text{sech}^2 x$ (no minus).
Inverse trig & hyperbolic derivatives (in MF19, but secure recall is faster): $\sqrt{1-x^2}$ , $1+x^2$ , $\sqrt{x^2+1}$ , $\sqrt{x^2-1}$ , $1-x^2$ denominators.
Parametric second derivative $\frac{d^2y}{dx^2} = \frac{d}{dt}\!\left(\frac{dy}{dx}\right)\div\frac{dx}{dt}$ .
Maclaurin series $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots$ — keep the factorials.
Useful identity $\text{sech}^2 x = 1 - \tanh^2 x$ for tidy Maclaurin differentiation.

How it’s examined

Section 2.3 is a Paper 2 (Further Pure Mathematics 2) staple, usually 6-10 marks. A typical question differentiates a composite hyperbolic or inverse function (chain/product/quotient), then asks for a second derivative from an implicit or parametric relation, or builds a few terms of a Maclaurin series by successive differentiation. Marks are lost through: inventing a minus sign on $\frac{d}{dx}\cosh x$ ; mixing the $\sqrt{x^2\pm 1}$ signs; forgetting the chain-rule factor; failing to divide by $\frac{dx}{dt}$ for the parametric second derivative; and dropping the factorial denominators in the series. Standard derivatives are in MF19, but Maclaurin coefficients must be derived in full — every derivative and evaluation earns method marks.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 2, Further Pure Mathematics 2); Cambridge List of Formulae (MF19) — differentiation table; 9231/21 & 9231/23 past papers and mark schemes. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Differentiation of Hyperbolic and Inverse Trigonometric Functions

Step-by-step solutions to past-paper-style questions on differentiation of hyperbolic and inverse trigonometric functions, written exactly the way a tutor would explain them at the board.

Question type:

1Differentiating a hyperbolic function with the chain rule
Getting startedDirect calculation• hyperbolic, chain-rule, easy
Question
Differentiate $y = \cosh(3x^2)$ with respect to $x$ .
Step-by-step solution
1. Step 1
  Identify the outer and inner functions. Outer is $\cosh(u)$ with $u = 3x^2$ . The standard derivative $\frac{d}{dx}\cosh u = \sinh u$ is in the MF19 table.
2. Step 2
  Apply the chain rule $\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}$ . Here $\frac{dy}{du} = \sinh u$ and $\frac{du}{dx} = 6x$ .
  $\frac{dy}{dx} = \sinh(3x^2)\cdot 6x$
3. Step 3
  Tidy up by writing the algebraic factor in front.
  $\frac{dy}{dx} = 6x\sinh(3x^2)$
Answer
$\dfrac{dy}{dx} = 6x\sinh(3x^2)$ .
Examiner tip
Watch the pairing: $\frac{d}{dx}\cosh x = \sinh x$ (NO minus sign), unlike $\frac{d}{dx}\cos x = -\sin x$ . The sign is the single biggest hyperbolic trap.
2Differentiating an inverse trig function with the chain rule
Getting startedDirect calculation• inverse-trig, chain-rule, easy
Question
Differentiate $y = \tan^{-1}(5x)$ with respect to $x$ .
Step-by-step solution
1. Step 1
  Recall the standard derivative (in MF19): $\frac{d}{dx}\tan^{-1}u = \frac{1}{1+u^2}\cdot\frac{du}{dx}$ . Here $u = 5x$ .
2. Step 2
  Differentiate the inner function: $\frac{du}{dx} = 5$ .
3. Step 3
  Apply the chain rule and substitute $u = 5x$ .
  $\frac{dy}{dx} = \frac{1}{1+(5x)^2}\cdot 5 = \frac{5}{1+25x^2}$
Answer
$\dfrac{dy}{dx} = \dfrac{5}{1+25x^2}$ .
Examiner tip
Square the WHOLE inner function: $(5x)^2 = 25x^2$ , not $5x^2$ . The $\frac{du}{dx}=5$ factor multiplies the standard result.
3Second derivative from an implicit relation
Building confidenceMulti-step problem• implicit, second-derivative, medium
Question
A curve is defined implicitly by $x^2 + y^2 = 25$ . Find $\dfrac{d^2y}{dx^2}$ in terms of $x$ and $y$ .
Step-by-step solution
1. Step 1
  Differentiate implicitly with respect to $x$ , remembering $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ .
  $2x + 2y\frac{dy}{dx} = 0 \;\Rightarrow\; \frac{dy}{dx} = -\frac{x}{y}$
2. Step 2
  Differentiate $\frac{dy}{dx} = -\frac{x}{y}$ again using the quotient rule, treating $y$ as a function of $x$ .
  $\frac{d^2y}{dx^2} = -\frac{(1)\cdot y - x\cdot\frac{dy}{dx}}{y^2}$
3. Step 3
  Substitute $\frac{dy}{dx} = -\frac{x}{y}$ into the numerator.
  $\frac{d^2y}{dx^2} = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2}$
4. Step 4
  Simplify by combining the numerator over $y$ and using $x^2 + y^2 = 25$ .
  $\frac{d^2y}{dx^2} = -\frac{y^2 + x^2}{y^3} = -\frac{25}{y^3}$
Answer
$\dfrac{d^2y}{dx^2} = -\dfrac{25}{y^3}$ .
Examiner tip
When differentiating $\frac{dy}{dx}$ again, the $x$ in the numerator forces a product/quotient rule and you MUST substitute the first derivative back in. Using the original equation to simplify $x^2+y^2$ earns the final A mark.
4Quotient rule with an inverse-sine factor
Building confidenceMulti-step problem• inverse-trig, quotient-rule, medium
Question
Differentiate $y = \dfrac{\sin^{-1}x}{x}$ with respect to $x$ , for $0 < x < 1$ .
Step-by-step solution
1. Step 1
  Identify $u$ and $v$ for the quotient rule $\frac{d}{dx}\frac{u}{v} = \frac{u'v - uv'}{v^2}$ . Let $u = \sin^{-1}x$ , $v = x$ .
2. Step 2
  Differentiate each part. From MF19, $u' = \frac{1}{\sqrt{1-x^2}}$ ; and $v' = 1$ .
  $u' = \frac{1}{\sqrt{1-x^2}}, \qquad v' = 1$
3. Step 3
  Assemble the quotient rule.
  $\frac{dy}{dx} = \frac{\frac{1}{\sqrt{1-x^2}}\cdot x - \sin^{-1}x\cdot 1}{x^2}$
4. Step 4
  Tidy the numerator.
  $\frac{dy}{dx} = \frac{x}{x^2\sqrt{1-x^2}} - \frac{\sin^{-1}x}{x^2} = \frac{1}{x\sqrt{1-x^2}} - \frac{\sin^{-1}x}{x^2}$
Answer
$\dfrac{dy}{dx} = \dfrac{1}{x\sqrt{1-x^2}} - \dfrac{\sin^{-1}x}{x^2}$ .
Examiner tip
Keep $\sin^{-1}x$ intact as a function — do not 'cancel' it with anything. The derivative $\frac{1}{\sqrt{1-x^2}}$ comes straight from MF19; quoting it correctly is the key M mark.
5Maclaurin series by successive differentiation
StretchMulti-step problem• maclaurin, successive-differentiation, hard
Question
Find the first three non-zero terms of the Maclaurin series for $f(x) = \tanh x$ , and state the range of validity.
Step-by-step solution
1. Step 1
  Differentiate successively. $f(x) = \tanh x$ , so $f'(x) = \text{sech}^2 x$ . Using $\text{sech}^2 x = 1 - \tanh^2 x$ keeps the algebra clean.
  $f'(x) = \text{sech}^2 x = 1 - \tanh^2 x$
2. Step 2
  Second derivative: differentiate $1 - \tanh^2 x$ via the chain rule.
  $f''(x) = -2\tanh x\,\text{sech}^2 x = -2\tanh x(1-\tanh^2 x)$
3. Step 3
  Third derivative: differentiate $-2\tanh x + 2\tanh^3 x$ .
  $f'''(x) = -2\text{sech}^2 x + 6\tanh^2 x\,\text{sech}^2 x$
4. Step 4
  Evaluate at $x = 0$ , using $\tanh 0 = 0$ and $\text{sech}\,0 = 1$ .
  $f(0) = 0,\;\; f'(0) = 1,\;\; f''(0) = 0,\;\; f'''(0) = -2$
5. Step 5
  Build the series $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$ .
  $\tanh x = 0 + (1)x + 0 + \frac{-2}{6}x^3 + \cdots = x - \frac{1}{3}x^3 + \cdots$
6. Step 6
  State validity. The expansion is valid for $|x| < \frac{\pi}{2}$ (the distance to the nearest singularity of $\tanh$ in the complex plane).
Answer
$\tanh x = x - \dfrac{1}{3}x^3 + \dfrac{2}{15}x^5 - \cdots$ ; first three non-zero terms $x - \dfrac{1}{3}x^3 + \dfrac{2}{15}x^5$ , valid for $|x| < \dfrac{\pi}{2}$ .
Examiner tip
Because consecutive coefficients vanish (odd function), you must differentiate further to reach three NON-ZERO terms — a common place to stop too early. Deriving each $f^{(n)}(0)$ explicitly is required: the series is not in MF19.
6Second derivative of a parametric relation
StretchMulti-step problem• parametric, second-derivative, hard
Question
A curve has parametric equations $x = \sinh t$ , $y = \cosh 2t$ . Find $\dfrac{d^2y}{dx^2}$ in terms of $t$ .
Step-by-step solution
1. Step 1
  Differentiate each equation w.r.t. $t$ . $\frac{dx}{dt} = \cosh t$ and $\frac{dy}{dt} = 2\sinh 2t$ (chain rule on $\cosh 2t$ ).
  $\frac{dx}{dt} = \cosh t, \qquad \frac{dy}{dt} = 2\sinh 2t$
2. Step 2
  First derivative $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ .
  $\frac{dy}{dx} = \frac{2\sinh 2t}{\cosh t}$
3. Step 3
  Simplify using $\sinh 2t = 2\sinh t\cosh t$ , so $\frac{2\sinh 2t}{\cosh t} = \frac{4\sinh t\cosh t}{\cosh t} = 4\sinh t$ .
  $\frac{dy}{dx} = 4\sinh t$
4. Step 4
  Second derivative: $\frac{d^2y}{dx^2} = \frac{d}{dt}\!\left(\frac{dy}{dx}\right)\div\frac{dx}{dt}$ . Here $\frac{d}{dt}(4\sinh t) = 4\cosh t$ .
  $\frac{d^2y}{dx^2} = \frac{4\cosh t}{\cosh t} = 4$
Answer
$\dfrac{d^2y}{dx^2} = 4$ .
Examiner tip
The classic parametric trap is dividing by $\frac{dx}{dt}$ a SECOND time when forming $\frac{d^2y}{dx^2}$ — you differentiate $\frac{dy}{dx}$ with respect to $t$ , then divide by $\frac{dx}{dt}$ once. Simplifying with $\sinh 2t = 2\sinh t\cosh t$ first makes the final division trivial.

Model Answers — Differentiation of Hyperbolic and Inverse Trigonometric Functions

High-scoring sample answers for differentiation of hyperbolic and inverse trigonometric functions on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
3 marks
Differentiate $y = \tanh(4x)$ with respect to $x$ .
Model answer
Standard derivative (MF19): $\frac{d}{dx}\tanh u = \text{sech}^2 u\cdot\frac{du}{dx}$ .

Here $u = 4x$ , so $\frac{du}{dx} = 4$ .

Apply the chain rule: $\frac{dy}{dx} = \text{sech}^2(4x)\cdot 4 = 4\,\text{sech}^2(4x)$
Why this scores
M1 for $\frac{d}{dx}\tanh u = \text{sech}^2 u$ , M1 for the chain-rule factor $\frac{du}{dx}=4$ , A1 for $4\,\text{sech}^2(4x)$ . Note the derivative of $\tanh$ is $\text{sech}^2$ , not $\sec^2$ .
Question 2
3 marks
Differentiate $y = \sinh^{-1}(x^2)$ with respect to $x$ .
Model answer
Standard derivative (MF19): $\frac{d}{dx}\sinh^{-1}u = \frac{1}{\sqrt{u^2+1}}\cdot\frac{du}{dx}$ .

Here $u = x^2$ , so $\frac{du}{dx} = 2x$ .

Substitute: $\frac{dy}{dx} = \frac{1}{\sqrt{(x^2)^2+1}}\cdot 2x = \frac{2x}{\sqrt{x^4+1}}$
Why this scores
M1 for the standard derivative with $+1$ under the root (it is $\sqrt{u^2+1}$ for $\sinh^{-1}$ , contrast $\sqrt{u^2-1}$ for $\cosh^{-1}$ ), M1 for $\frac{du}{dx}=2x$ , A1 for the simplified answer.
Question 3
4 marks
Differentiate $y = x^2\cosh^{-1}x$ with respect to $x$ , for $x > 1$ .
Model answer
Product rule $\frac{d}{dx}(uv) = u'v + uv'$ with $u = x^2$ , $v = \cosh^{-1}x$ .

Derivatives: $u' = 2x$ and, from MF19, $v' = \frac{1}{\sqrt{x^2-1}}$ .

Assemble: $\frac{dy}{dx} = 2x\cosh^{-1}x + x^2\cdot\frac{1}{\sqrt{x^2-1}}$

$\frac{dy}{dx} = 2x\cosh^{-1}x + \frac{x^2}{\sqrt{x^2-1}}$
Why this scores
M1 for correct product-rule structure, B1 for $\frac{d}{dx}\cosh^{-1}x = \frac{1}{\sqrt{x^2-1}}$ (the $-1$ under the root is essential), M1 for $\frac{d}{dx}x^2 = 2x$ , A1 for the fully assembled derivative. Leave $\cosh^{-1}x$ as-is.
Question 4
5 marks
A curve is given by $y = \tan^{-1}(2x)$ . (a) Find $\dfrac{dy}{dx}$ . (b) Hence find $\dfrac{d^2y}{dx^2}$ .
Model answer
(a) Using $\frac{d}{dx}\tan^{-1}u = \frac{1}{1+u^2}\cdot\frac{du}{dx}$ with $u = 2x$ , $\frac{du}{dx}=2$ : $\frac{dy}{dx} = \frac{2}{1+4x^2}$

(b) Write $\frac{dy}{dx} = 2(1+4x^2)^{-1}$ and differentiate using the chain rule: $\frac{d^2y}{dx^2} = 2\cdot(-1)(1+4x^2)^{-2}\cdot 8x = -\frac{16x}{(1+4x^2)^2}$
Why this scores
Part (a): M1 standard derivative + chain factor, A1 for $\frac{2}{1+4x^2}$ . Part (b): M1 for rewriting as a power and applying the chain rule, A1 A1 for the correct numerator $-16x$ and squared denominator. The inner derivative $\frac{d}{dx}(1+4x^2)=8x$ is where marks are lost.
Question 5
7 marks
Show that the Maclaurin series for $f(x) = \ln(\cosh x)$ begins $\dfrac{1}{2}x^2 - \dfrac{1}{12}x^4 + \cdots$ , by successive differentiation.
Model answer
Differentiate successively. With $f(x) = \ln(\cosh x)$ : $f'(x) = \frac{\sinh x}{\cosh x} = \tanh x$

$f''(x) = \text{sech}^2 x = 1 - \tanh^2 x$

$f'''(x) = -2\tanh x\,\text{sech}^2 x$

$f^{(4)}(x) = -2\,\text{sech}^4 x + 4\tanh^2 x\,\text{sech}^2 x$

Evaluate at $x = 0$ (using $\tanh 0 = 0$ , $\text{sech}\,0 = 1$ , $\cosh 0 = 1$ ): $f(0) = \ln 1 = 0,\quad f'(0) = 0,\quad f''(0) = 1,\quad f'''(0) = 0,\quad f^{(4)}(0) = -2$

Build the series: $f(x) = 0 + 0\cdot x + \frac{1}{2!}x^2 + \frac{0}{3!}x^3 + \frac{-2}{4!}x^4 + \cdots$

$f(x) = \frac{1}{2}x^2 - \frac{2}{24}x^4 + \cdots = \frac{1}{2}x^2 - \frac{1}{12}x^4 + \cdots \quad\text{(as required)}$
Why this scores
M1 for $f'(x)=\tanh x$ , M1 for $f''(x)=\text{sech}^2 x$ , M1 for a correct higher derivative, A1 for all values at $0$ , M1 for the Maclaurin construction with factorials, A1 A1 for the two stated coefficients. As a 'Show that', every derivative and evaluation must be shown — quoting the answer scores nothing.
Question 6
7 marks
A curve has parametric equations $x = \tan^{-1}t$ , $y = \ln(1+t^2)$ . Find $\dfrac{d^2y}{dx^2}$ in terms of $t$ .
Model answer
Differentiate w.r.t. $t$ : $\frac{dx}{dt} = \frac{1}{1+t^2}, \qquad \frac{dy}{dt} = \frac{2t}{1+t^2}$

First derivative: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t/(1+t^2)}{1/(1+t^2)} = 2t$

Second derivative: differentiate $\frac{dy}{dx}=2t$ w.r.t. $t$ , then divide by $\frac{dx}{dt}$ : $\frac{d}{dt}\!\left(\frac{dy}{dx}\right) = 2$

$\frac{d^2y}{dx^2} = \frac{2}{\,1/(1+t^2)\,} = 2(1+t^2)$
Why this scores
M1 A1 for $\frac{dx}{dt}$ and $\frac{dy}{dt}$ , M1 A1 for $\frac{dy}{dx}=2t$ after cancelling $(1+t^2)$ , M1 for differentiating $\frac{dy}{dx}$ w.r.t. $t$ then dividing by $\frac{dx}{dt}$ , A1 for $2(1+t^2)$ . The fatal error is forgetting that $\frac{d^2y}{dx^2}$ requires dividing by $\frac{dx}{dt}$ , not multiplying.

Key Formulae — Differentiation of Hyperbolic and Inverse Trigonometric Functions

The formulae you need to memorise for differentiation of hyperbolic and inverse trigonometric functions on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Derivatives of hyperbolic functions
$\frac{d}{dx}\sinh x = \cosh x,\quad \frac{d}{dx}\cosh x = \sinh x,\quad \frac{d}{dx}\tanh x = \text{sech}^2 x$
$\sinh x$
hyperbolic sine, $\frac{e^x - e^{-x}}{2}$
$\cosh x$
hyperbolic cosine, $\frac{e^x + e^{-x}}{2}$
$\text{sech}\,x$
$\frac{1}{\cosh x}$ , the hyperbolic secant
When to use
Differentiating $\sinh$ , $\cosh$ , $\tanh$ (and via the quotient rule $\text{sech}$ , $\text{cosech}$ , $\coth$ ). No minus sign on $\frac{d}{dx}\cosh x$ . In MF19.
Derivatives of inverse trigonometric functions
$\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}},\qquad \frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$
$\sin^{-1}x$
inverse sine (arcsin), valid $-1 < x < 1$
$\tan^{-1}x$
inverse tangent (arctan), valid for all $x$
When to use
Differentiating inverse trig functions; combine with the chain rule for $\sin^{-1}u$ , $\tan^{-1}u$ . In MF19.
Derivatives of inverse hyperbolic functions
$\frac{d}{dx}\sinh^{-1}x = \frac{1}{\sqrt{x^2+1}},\quad \frac{d}{dx}\cosh^{-1}x = \frac{1}{\sqrt{x^2-1}},\quad \frac{d}{dx}\tanh^{-1}x = \frac{1}{1-x^2}$
$\sinh^{-1}x$
inverse hyperbolic sine, valid for all $x$
$\cosh^{-1}x$
inverse hyperbolic cosine, valid $x \geq 1$
$\tanh^{-1}x$
inverse hyperbolic tangent, valid $|x| < 1$
When to use
Differentiating inverse hyperbolic functions. Note the sign under the root: $+1$ for $\sinh^{-1}$ , $-1$ for $\cosh^{-1}$ . In MF19.
Second derivative of a parametric curve
$\frac{d^2y}{dx^2} = \frac{d}{dt}\!\left(\frac{dy}{dx}\right)\div\frac{dx}{dt}$
$\frac{dy}{dx}$
first derivative, $= \frac{dy/dt}{dx/dt}$
$t$
the parameter
When to use
Finding $\frac{d^2y}{dx^2}$ for $x = x(t)$ , $y = y(t)$ . Differentiate $\frac{dy}{dx}$ w.r.t. $t$ , then divide by $\frac{dx}{dt}$ ONCE.
Maclaurin series
$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$
$f^{(n)}(0)$
the $n$ -th derivative evaluated at $x = 0$
$n!$
factorial, $n! = n(n-1)\cdots 2\cdot 1$
When to use
Building a series by successive differentiation: compute $f(0), f'(0), f''(0), \ldots$ then substitute. Derive the coefficients — only a few standard series are in MF19.

Key Definitions and Keywords — Differentiation of Hyperbolic and Inverse Trigonometric Functions

Definitions to memorise and the exact keywords mark schemes credit for differentiation of hyperbolic and inverse trigonometric functions answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Hyperbolic function
Examiner keyword
A function defined from exponentials: $\sinh x = \frac{e^x - e^{-x}}{2}$ , $\cosh x = \frac{e^x + e^{-x}}{2}$ , $\tanh x = \frac{\sinh x}{\cosh x}$ . Their derivatives follow the pattern $\frac{d}{dx}\sinh x = \cosh x$ , $\frac{d}{dx}\cosh x = \sinh x$ .
Inverse trigonometric function
Examiner keyword
The inverse of a (restricted) trig function, e.g. $\sin^{-1}x$ returns the angle whose sine is $x$ . Standard derivatives such as $\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}$ appear in the MF19 differentiation table.
Inverse hyperbolic function
Examiner keyword
The inverse of a hyperbolic function, e.g. $\sinh^{-1}x$ . Its derivative is $\frac{1}{\sqrt{x^2+1}}$ ; for $\cosh^{-1}x$ it is $\frac{1}{\sqrt{x^2-1}}$ . Both are in MF19.
Maclaurin series
Examiner keyword
A power-series expansion of $f(x)$ about $x = 0$ : $f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$ . Coefficients are found by successive differentiation and evaluation at $0$ .
Range of validity
Examiner keyword
The set of $x$ -values for which a Maclaurin (power) series converges to the function, e.g. $|x| < 1$ for $\ln(1+x)$ . Candidates must state it where relevant.
Successive differentiation
Repeatedly differentiating a function to obtain $f'(x), f''(x), f'''(x), \ldots$ , used to generate Maclaurin-series coefficients $f^{(n)}(0)$ .

Common Mistakes and Misconceptions — Differentiation of Hyperbolic and Inverse Trigonometric Functions

The traps other students keep falling into on differentiation of hyperbolic and inverse trigonometric functions questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Writing $\frac{d}{dx}\cosh x = -\sinh x$ (inventing a minus sign)
9231 Paper 2 Examiner Reports — recurring comment
Why it happens
Carrying over the circular result $\frac{d}{dx}\cos x = -\sin x$ to the hyperbolic case by analogy.
How to avoid it
Memorise: BOTH $\frac{d}{dx}\sinh x = \cosh x$ and $\frac{d}{dx}\cosh x = \sinh x$ are positive. The sign change only happens for circular functions.
✕Confusing $\sqrt{x^2+1}$ ( $\sinh^{-1}$ ) with $\sqrt{x^2-1}$ ( $\cosh^{-1}$ ) under the root
9231 Paper 2 mark schemes
Why it happens
The two inverse-hyperbolic derivatives look almost identical; the $\pm 1$ is easy to misremember.
How to avoid it
Link it to domains: $\cosh^{-1}x$ needs $x \geq 1$ , so $\sqrt{x^2-1}$ (real only there); $\sinh^{-1}x$ is defined for all $x$ , so $\sqrt{x^2+1}$ (always real). Or just read it off MF19.
✕Omitting the $\frac{du}{dx}$ chain-rule factor, e.g. $\frac{d}{dx}\tan^{-1}(5x) = \frac{1}{1+25x^2}$
9231 Paper 2 Examiner Reports
Why it happens
Quoting the bare standard derivative and forgetting the inner function must be differentiated too.
How to avoid it
Always write the standard derivative with $u$ , then multiply by $\frac{du}{dx}$ . For $\tan^{-1}(5x)$ the answer is $\frac{5}{1+25x^2}$ .
✕Forming $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$ w.r.t. $t$ WITHOUT dividing by $\frac{dx}{dt}$
9231 Paper 2 — parametric questions
Why it happens
Treating the parameter $t$ as if it were $x$ ; forgetting the extra chain-rule step.
How to avoid it
Use $\frac{d^2y}{dx^2} = \frac{d}{dt}\!\left(\frac{dy}{dx}\right)\div\frac{dx}{dt}$ . Differentiate, THEN divide by $\frac{dx}{dt}$ once.
✕Dropping the factorial denominators when assembling the Maclaurin series
9231 Paper 2 Examiner Reports
Why it happens
Writing $f(0) + f'(0)x + f''(0)x^2 + \cdots$ without the $\frac{1}{2!}, \frac{1}{3!}, \ldots$ factors.
How to avoid it
The coefficient of $x^n$ is $\frac{f^{(n)}(0)}{n!}$ . Write the full template with factorials before substituting derivative values.
✕Stopping at the first 'three terms' when some coefficients are zero (e.g. for an odd function)
9231 Paper 2 — series questions
Why it happens
A request for three NON-ZERO terms is read as three terms, so the student stops after vanishing coefficients.
How to avoid it
Read the demand carefully: keep differentiating until you have the required number of NON-ZERO terms. Odd/even functions have alternating zero coefficients.

Differentiation of Hyperbolic and Inverse Trigonometric Functions

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Differentiating a hyperbolic function with the chain rule

2Differentiating an inverse trig function with the chain rule

3Second derivative from an implicit relation

4Quotient rule with an inverse-sine factor

5Maclaurin series by successive differentiation

6Second derivative of a parametric relation

Derivatives of hyperbolic functions

Derivatives of inverse trigonometric functions

Derivatives of inverse hyperbolic functions

Second derivative of a parametric curve

Maclaurin series

Hyperbolic function

Inverse trigonometric function

Inverse hyperbolic function

Maclaurin series

Range of validity

Successive differentiation

✕Writing ddxcosh⁡x=−sinh⁡x\frac{d}{dx}\cosh x = -\sinh xdxd​coshx=−sinhx (inventing a minus sign)

✕Confusing x2+1\sqrt{x^2+1}x2+1​ (sinh⁡−1\sinh^{-1}sinh−1) with x2−1\sqrt{x^2-1}x2−1​ (cosh⁡−1\cosh^{-1}cosh−1) under the root

✕Omitting the dudx\frac{du}{dx}dxdu​ chain-rule factor, e.g. ddxtan⁡−1(5x)=11+25x2\frac{d}{dx}\tan^{-1}(5x) = \frac{1}{1+25x^2}dxd​tan−1(5x)=1+25x21​

✕Forming d2ydx2\frac{d^2y}{dx^2}dx2d2y​ by differentiating dydx\frac{dy}{dx}dxdy​ w.r.t. ttt WITHOUT dividing by dxdt\frac{dx}{dt}dtdx​

✕Dropping the factorial denominators when assembling the Maclaurin series

✕Stopping at the first 'three terms' when some coefficients are zero (e.g. for an odd function)

✕Writing $\frac{d}{dx}\cosh x = -\sinh x$ (inventing a minus sign)

✕Confusing $\sqrt{x^2+1}$ ( $\sinh^{-1}$ ) with $\sqrt{x^2-1}$ ( $\cosh^{-1}$ ) under the root

✕Omitting the $\frac{du}{dx}$ chain-rule factor, e.g. $\frac{d}{dx}\tan^{-1}(5x) = \frac{1}{1+25x^2}$

✕Forming $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$ w.r.t. $t$ WITHOUT dividing by $\frac{dx}{dt}$