Differentiating Hyperbolic Functions

Differentiating the hyperbolic functions is one of the cleanest results in the whole calculus syllabus, because it flows straight out of the exponential definitions: $\sinh$sinh and $\cosh$cosh differentiate into each other with no sign change, the single most important contrast with the circular functions, where $\cos$cos picks up a minus sign. This lesson derives every standard derivative from $\tfrac{\mathrm d}{\mathrm dx}e^x=e^x$dxd​ex=ex, extends them by the chain, product and quotient rules, and — crucially for the integration lesson that follows — differentiates the inverse hyperbolic functions to produce the standard integrals $\int\tfrac{1}{\sqrt{x^2+1}}=\operatorname{arsinh} x$∫x2+1​1​=arsinhx and friends.

---

1 Spec mapping

Hyperbolic differentiation is compulsory pure content for Papers 1 and 2 of AQA Further Mathematics (7367). Applying the standard derivatives with the chain, product and quotient rules is AO1; proving the derivatives from the definitions, and the implicit-differentiation derivations of the inverse derivatives, are AO2. It rests entirely on the differentiation toolkit of A-Level Maths (chain/product/quotient rules, differentiating $e^{kx}$ekx) and on the previous two lessons' definitions and identities. The inverse derivatives are the linchpin: every standard integral in the next lesson is one of these derivatives run backwards, so the two topics are really one. In practice, examiners assess this content both directly ("differentiate the following") and inside larger questions — finding the gradient of a tangent to a hyperbola, locating a stationary point of a catenary-type function, or supplying the antiderivative needed for a definite integral — so fluency here pays dividends right across the pure papers.

---

2 Core theory — the three standard derivatives

$\boxed{\ \frac{\mathrm d}{\mathrm dx}\sinh x=\cosh x,\qquad \frac{\mathrm d}{\mathrm dx}\cosh x=\sinh x,\qquad \frac{\mathrm d}{\mathrm dx}\tanh x=\operatorname{sech}^2 x\ }$ dxd​sinhx=coshx,dxd​coshx=sinhx,dxd​tanhx=sech2x ​

Compare the circular derivatives: $\tfrac{\mathrm d}{\mathrm dx}\sin x=\cos x$dxd​sinx=cosx, $\tfrac{\mathrm d}{\mathrm dx}\cos x=-\sin x$dxd​cosx=−sinx, $\tfrac{\mathrm d}{\mathrm dx}\tan x=\sec^2 x$dxd​tanx=sec2x. The one structural difference is the missing minus sign in $\tfrac{\mathrm d}{\mathrm dx}\cosh x=\sinh x$dxd​coshx=sinhx (it is $+\sinh x$+sinhx, not $-\sinh x$−sinhx). Everything else looks identical with hyperbolic names. This single sign is the root of every later difference — for instance, why $\int\operatorname{sech}^2=\tanh$∫sech2=tanh matches trig, but the inverse-function integrals split into $\operatorname{arsinh}/\operatorname{arcosh}$arsinh/arcosh rather than a single $\arcsin$arcsin.

Proofs from the definitions. Each is a one-line application of $\tfrac{\mathrm d}{\mathrm dx}e^{x}=e^{x}$dxd​ex=ex and $\tfrac{\mathrm d}{\mathrm dx}e^{-x}=-e^{-x}$dxd​e−x=−e−x:

$\frac{\mathrm d}{\mathrm dx}\sinh x=\frac{\mathrm d}{\mathrm dx}\!\left(\frac{e^x-e^{-x}}{2}\right)=\frac{e^x+e^{-x}}{2}=\cosh x,$dxd​sinhx=dxd​(2ex−e−x​)=2ex+e−x​=coshx,

$\frac{\mathrm d}{\mathrm dx}\cosh x=\frac{\mathrm d}{\mathrm dx}\!\left(\frac{e^x+e^{-x}}{2}\right)=\frac{e^x-e^{-x}}{2}=\sinh x.$dxd​coshx=dxd​(2ex+e−x​)=2ex−e−x​=sinhx.

For $\tanh$tanh, apply the quotient rule to $\dfrac{\sinh x}{\cosh x}$coshxsinhx​ and use $\cosh^2 x-\sinh^2 x=1$cosh2x−sinh2x=1:

$\frac{\mathrm d}{\mathrm dx}\tanh x=\frac{\cosh x\cdot\cosh x-\sinh x\cdot\sinh x}{\cosh^2 x}=\frac{\cosh^2 x-\sinh^2 x}{\cosh^2 x}=\frac{1}{\cosh^2 x}=\operatorname{sech}^2 x.$dxd​tanhx=cosh2xcoshx⋅coshx−sinhx⋅sinhx​=cosh2xcosh2x−sinh2x​=cosh2x1​=sech2x.

Notice that the positive derivative of $\cosh$cosh is exactly what makes the $\tanh$tanh numerator come out as $+1$+1 (via $\cosh^2-\sinh^2$cosh2−sinh2); the trig analogue gets $+1$+1 from $\cos^2+\sin^2$cos2+sin2 instead. Same destination, different route.

There is a deeper way to see why no minus sign appears, and it is worth holding onto because it explains the entire topic in one stroke. Differentiation of $e^{x}$ex leaves it unchanged, and $\sinh$sinh, $\cosh$cosh are built purely from $e^{x}$ex and $e^{-x}$e−x with real coefficients. The circular functions, by contrast, are built from $e^{\mathrm ix}$eix and $e^{-\mathrm ix}$e−ix, and each differentiation pulls down a factor of $\mathrm i$i; two such factors give $\mathrm i^2=-1$i2=−1, which is exactly the minus sign that appears in $\tfrac{\mathrm d^2}{\mathrm dx^2}\cos x=-\cos x$dx2d2​cosx=−cosx. The hyperbolic functions never see that $\mathrm i$i, so they obey $\tfrac{\mathrm d^2}{\mathrm dx^2}\cosh x=+\cosh x$dx2d2​coshx=+coshx. In short, the missing minus sign is not an arbitrary fact to memorise — it is the absence of the imaginary unit, and it propagates consistently through every derivative, every integral, and every differential equation in the strand. Anchoring the whole topic to this one idea is far more robust than trying to remember a table of signs.

---

3 Reciprocal-function derivatives

$\frac{\mathrm d}{\mathrm dx}\operatorname{sech} x=-\operatorname{sech} x\tanh x,\qquad \frac{\mathrm d}{\mathrm dx}\operatorname{cosech} x=-\operatorname{cosech} x\coth x,\qquad \frac{\mathrm d}{\mathrm dx}\coth x=-\operatorname{cosech}^2 x.$dxd​sechx=−sechxtanhx,dxd​cosechx=−cosechxcothx,dxd​cothx=−cosech2x.

These all carry a minus sign (note: the trig cousins $\tfrac{\mathrm d}{\mathrm dx}\sec x=\sec x\tan x$dxd​secx=secxtanx and $\tfrac{\mathrm d}{\mathrm dx}\operatorname{cosec} x=-\operatorname{cosec} x\cot x$dxd​cosecx=−cosecxcotx split between plus and minus, whereas all three hyperbolic reciprocal derivatives are negative). You do not need to memorise these three separately — each follows in two lines from the standard derivatives plus the chain or quotient rule, and that is exactly how examiners expect you to reproduce them if asked. For example, writing $\operatorname{sech} x=(\cosh x)^{-1}$sechx=(coshx)−1 and using the chain rule:

$\frac{\mathrm d}{\mathrm dx}(\cosh x)^{-1}=-(\cosh x)^{-2}\cdot\sinh x=-\frac{1}{\cosh x}\cdot\frac{\sinh x}{\cosh x}=-\operatorname{sech} x\tanh x.$dxd​(coshx)−1=−(coshx)−2⋅sinhx=−coshx1​⋅coshxsinhx​=−sechxtanhx.

And $\coth x=\dfrac{\cosh x}{\sinh x}$cothx=sinhxcoshx​ by the quotient rule gives $\dfrac{\sinh^2 x-\cosh^2 x}{\sinh^2 x}=\dfrac{-1}{\sinh^2 x}=-\operatorname{cosech}^2 x$sinh2xsinh2x−cosh2x​=sinh2x−1​=−cosech2x.

---

4 The chain, product and quotient rules

The standard derivatives combine with the usual rules exactly as the circular ones do — there is nothing new to learn about the rules themselves, only the three elementary derivatives feeding into them. This is liberating: every product-rule, quotient-rule and chain-rule technique you drilled at A-Level transfers verbatim, so a question like "differentiate $x^3\tanh(2x)$x3tanh(2x)" is solved by the same mechanical process as "differentiate $x^3\tan(2x)$x3tan(2x)", merely substituting $\operatorname{sech}^2$sech2 for $\sec^2$sec2. With the chain rule:

$\frac{\mathrm d}{\mathrm dx}\sinh f(x)=f'(x)\cosh f(x),\quad \frac{\mathrm d}{\mathrm dx}\cosh f(x)=f'(x)\sinh f(x),\quad \frac{\mathrm d}{\mathrm dx}\tanh f(x)=f'(x)\operatorname{sech}^2 f(x).$dxd​sinhf(x)=f′(x)coshf(x),dxd​coshf(x)=f′(x)sinhf(x),dxd​tanhf(x)=f′(x)sech2f(x).

Worked Example 1 — chain rule

Differentiate $y=\sinh(3x^2)$y=sinh(3x2).

With $f(x)=3x^2$f(x)=3x2, $f'(x)=6x$f′(x)=6x:

$\frac{\mathrm dy}{\mathrm dx}=6x\cosh(3x^2).$dxdy​=6xcosh(3x2).

Mark scheme: M1 chain rule with outer derivative $\cosh$cosh; A1 $6x\cosh(3x^2)$6xcosh(3x2) (inner derivative $6x$6x correct).

Worked Example 2 — recognising a double-angle simplification

Differentiate $y=\cosh^2 x$y=cosh2x.

$\frac{\mathrm dy}{\mathrm dx}=2\cosh x\cdot\sinh x=\sinh 2x \quad(\text{using }2\sinh x\cosh x=\sinh 2x).$dxdy​=2coshx⋅sinhx=sinh2x(using 2sinhxcoshx=sinh2x).

Mark scheme: M1 chain rule on $(\cosh x)^2$(coshx)2; A1 $2\sinh x\cosh x$2sinhxcoshx; A1 simplify to $\sinh 2x$sinh2x.

Worked Example 3 — product rule

Differentiate $y=x^2\tanh x$y=x2tanhx.

$\frac{\mathrm dy}{\mathrm dx}=\underbrace{2x}_{u'}\,\tanh x+x^2\,\underbrace{\operatorname{sech}^2 x}_{v'}=2x\tanh x+x^2\operatorname{sech}^2 x.$dxdy​=u′2x​​tanhx+x2v′sech2x​​=2xtanhx+x2sech2x.

Mark scheme: M1 product rule; A1 $2x\tanh x$2xtanhx; A1 $x^2\operatorname{sech}^2 x$x2sech2x.

Worked Example 4 — product rule with a chain

Differentiate $y=e^x\cosh 2x$y=excosh2x.

$\frac{\mathrm dy}{\mathrm dx}=e^x\cosh 2x+e^x\cdot 2\sinh 2x=e^x\big(\cosh 2x+2\sinh 2x\big).$dxdy​=excosh2x+ex⋅2sinh2x=ex(cosh2x+2sinh2x).

Mark scheme: M1 product rule; M1 chain rule on $\cosh 2x$cosh2x giving $2\sinh 2x$2sinh2x; A1 factor to $e^x(\cosh 2x+2\sinh 2x)$ex(cosh2x+2sinh2x).

Worked Example 4b — quotient rule

Differentiate $y=\dfrac{\sinh x}{x}$y=xsinhx​ for $x\ne0$x=0.

By the quotient rule with $u=\sinh x$u=sinhx (so $u'=\cosh x$u′=coshx) and $v=x$v=x (so $v'=1$v′=1):

$\frac{\mathrm dy}{\mathrm dx}=\frac{x\cosh x-\sinh x}{x^2}.$dxdy​=x2xcoshx−sinhx​.

Mark scheme: M1 quotient rule, correct structure; A1 numerator $x\cosh x-\sinh x$xcoshx−sinhx; A1 divide by $x^2$x2. A useful check on sign: near $x=0$x=0, $\sinh x\approx x$sinhx≈x so the numerator $x\cosh x-\sinh x\approx x-x=0$xcoshx−sinhx≈x−x=0, confirming the function is smooth through the (removable) point at the origin where $\sinh x/x\to1$sinhx/x→1.

---

5 Derivatives of the inverse hyperbolic functions

These three results are the engine of the integration lesson:

$\boxed{\ \frac{\mathrm d}{\mathrm dx}\operatorname{arsinh} x=\frac{1}{\sqrt{x^2+1}},\quad \frac{\mathrm d}{\mathrm dx}\operatorname{arcosh} x=\frac{1}{\sqrt{x^2-1}}\ (x>1),\quad \frac{\mathrm d}{\mathrm dx}\operatorname{artanh} x=\frac{1}{1-x^2}\ (|x|<1)\ }$ dxd​arsinhx=x2+1​1​,dxd​arcoshx=x2−1​1​ (x>1),dxd​artanhx=1−x21​ (∣x∣<1) ​

Proof ($\operatorname{arsinh}$arsinh) by implicit differentiation. Let $y=\operatorname{arsinh} x$y=arsinhx, so $\sinh y=x$sinhy=x. Differentiate both sides with respect to $x$x:

$\cosh y\,\frac{\mathrm dy}{\mathrm dx}=1\ \implies\ \frac{\mathrm dy}{\mathrm dx}=\frac{1}{\cosh y}=\frac{1}{\sqrt{1+\sinh^2 y}}=\frac{1}{\sqrt{1+x^2}}.$coshydxdy​=1 ⟹ dxdy​=coshy1​=1+sinh2y​1​=1+x2​1​.

The step $\cosh y=\sqrt{1+\sinh^2 y}$coshy=1+sinh2y​ takes the positive root, valid because $\cosh y\ge1>0$coshy≥1>0. The $\operatorname{arcosh}$arcosh proof is identical with $\cosh y=x$coshy=x and $\sinh y=\sqrt{\cosh^2 y-1}=\sqrt{x^2-1}$sinhy=cosh2y−1​=x2−1​ (positive on the branch $y>0$y>0), giving $\tfrac{1}{\sqrt{x^2-1}}$x2−1​1​. For $\operatorname{artanh}$artanh, $\tanh y=x$tanhy=x gives $\operatorname{sech}^2 y\,\tfrac{\mathrm dy}{\mathrm dx}=1$sech2ydxdy​=1, so $\tfrac{\mathrm dy}{\mathrm dx}=\tfrac{1}{\operatorname{sech}^2 y}=\tfrac{1}{1-\tanh^2 y}=\tfrac{1}{1-x^2}$dxdy​=sech2y1​=1−tanh2y1​=1−x21​. These mirror the inverse-trig derivatives $\tfrac{\mathrm d}{\mathrm dx}\arcsin x=\tfrac{1}{\sqrt{1-x^2}}$dxd​arcsinx=1−x2​1​ and $\tfrac{\mathrm d}{\mathrm dx}\arctan x=\tfrac{1}{1+x^2}$dxd​arctanx=1+x21​ — once again the sign under the root is the only difference.

Why does implicit differentiation work so cleanly here? The trick is that we never need the explicit logarithmic form of the inverse function at all. We know $\sinh y=x$sinhy=x by definition, and differentiating that equation — not the messy $y=\ln(x+\sqrt{x^2+1})$y=ln(x+x2+1​) — gives the derivative almost instantly, because the derivative of $\sinh y$sinhy is the friendly $\cosh y$coshy. The only real work is re-expressing the answer $\tfrac{1}{\cosh y}$coshy1​ back in terms of $x$x, which is where the fundamental identity earns its keep: $\cosh y=\sqrt{1+\sinh^2 y}=\sqrt{1+x^2}$coshy=1+sinh2y​=1+x2​. This is a general and powerful technique — it is exactly how one differentiates any inverse function (including $\arcsin$arcsin, $\arctan$arctan, and even $\ln$ln itself, viewed as the inverse of $e^x$ex) without ever inverting the formula explicitly. If you can reproduce this three-line argument, you can reconstruct all three inverse-hyperbolic derivatives under exam pressure, even if you have forgotten which sign goes under which root.

You can also sanity-check each inverse derivative by its sign and domain. $\tfrac{1}{\sqrt{x^2+1}}$x2+1​1​ is always positive and defined for all $x$x — matching the fact that $\operatorname{arsinh}$arsinh is everywhere increasing. $\tfrac{1}{\sqrt{x^2-1}}$x2−1​1​ is positive but defined only for $x>1$x>1, and blows up as $x\to1^+$x→1+ — matching $\operatorname{arcosh}$arcosh's vertical tangent at $(1,0)$(1,0). And $\tfrac{1}{1-x^2}$1−x21​ is positive on $(-1,1)$(−1,1) and diverges at $x=\pm1$x=±1 — matching $\operatorname{artanh}$artanh's vertical asymptotes. A derivative whose sign or domain disagrees with the graph is a red flag for an algebra error.

Worked Example 5 — inverse with a chain

Differentiate $y=\operatorname{arsinh}(2x)$y=arsinh(2x).

$\frac{\mathrm dy}{\mathrm dx}=\frac{1}{\sqrt{(2x)^2+1}}\cdot 2=\frac{2}{\sqrt{4x^2+1}}.$dxdy​=(2x)2+1​1​⋅2=4x2+1​2​.

Mark scheme: M1 $\tfrac{\mathrm d}{\mathrm dx}\operatorname{arsinh} u=\tfrac{u'}{\sqrt{u^2+1}}$dxd​arsinhu=u2+1​u′​; A1 $\tfrac{2}{\sqrt{4x^2+1}}$4x2+1​2​.

---

6 Worked examples — parametric and harder cases

Worked Example 6 — parametric differentiation

A curve is given by $x=2\cosh t,\ y=3\sinh t$x=2cosht, y=3sinht. Find $\dfrac{\mathrm dy}{\mathrm dx}$dxdy​ in terms of $t$t.

$\frac{\mathrm dx}{\mathrm dt}=2\sinh t,\qquad \frac{\mathrm dy}{\mathrm dt}=3\cosh t,$dtdx​=2sinht,dtdy​=3cosht,

$\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt}=\frac{3\cosh t}{2\sinh t}=\tfrac32\coth t.$dxdy​=dx/dtdy/dt​=2sinht3cosht​=23​cotht.

Mark scheme: M1 $\tfrac{\mathrm dy}{\mathrm dx}=\tfrac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt}$dxdy​=dx/dtdy/dt​; A1 both component derivatives; A1 $\tfrac32\coth t$23​cotht. (This curve is the hyperbola $\tfrac{x^2}{4}-\tfrac{y^2}{9}=1$4x2​−9y2​=1, since $\cosh^2 t-\sinh^2 t=1$cosh2t−sinh2t=1 — a neat synoptic check.)

Worked Example 7 — a stationary point

Find the stationary point of $y=\cosh x-x$y=coshx−x and determine its nature.

$\frac{\mathrm dy}{\mathrm dx}=\sinh x-1=0\ \implies\ \sinh x=1\ \implies\ x=\operatorname{arsinh}1=\ln(1+\sqrt2).$dxdy​=sinhx−1=0 ⟹ sinhx=1 ⟹ x=arsinh1=ln(1+2​).