Unit 3, lesson 32

Derivatives of Hyperbolic Functions

What sinh, cosh and tanh are, why their derivatives look like trigonometric ones with different signs, and how to differentiate their inverses on the right real domains.

  • Theory part 63
  • Practice part 64
  • Derivatives of Elementary Functions

Theory, part 63

Practice, part 64

Hold a chain or a cable loosely by its two ends and let it hang. The curve it settles into looks like a parabola but is not one. It is a scaled copy of the graph of the hyperbolic cosine. The hyperbolic functions are built from exe^x and e−xe^{-x}, and they behave so much like sine and cosine that their formulas are easy to remember once you know where the signs differ.

Key terms

Exponential definitions

Definition For every real xx, the hyperbolic sine and hyperbolic cosine are

sinh⁡x=ex−e−x2,cosh⁡x=ex+e−x2.\sinh x = \frac{e^x - e^{-x}}{2}, \qquad \cosh x = \frac{e^x + e^{-x}}{2}.

In plain words The function cosh⁡x\cosh x is the even part of exe^x and sinh⁡x\sinh x is its odd part. Adding them gives back exe^x.

Example At x=ln⁡2x = \ln 2 we have ex=2e^x = 2 and e−x=12e^{-x} = \frac{1}{2}, so

sinh⁡(ln⁡2)=2−1/22=34,cosh⁡(ln⁡2)=2+1/22=54.\sinh(\ln 2) = \frac{2 - 1/2}{2} = \frac{3}{4}, \qquad \cosh(\ln 2) = \frac{2 + 1/2}{2} = \frac{5}{4}.

Hyperbolic sine

Definition The function sinh⁡\sinh is odd, defined and differentiable on R\mathbb{R}, with range R\mathbb{R}, and

ddxsinh⁡x=ex+e−x2=cosh⁡x.\frac{d}{dx} \sinh x = \frac{e^x + e^{-x}}{2} = \cosh x.

In plain words The slope of sinh⁡\sinh is cosh⁡\cosh, which is always at least 1, so sinh⁡\sinh is increasing everywhere.

Example At x=0x = 0 the value is sinh⁡0=0\sinh 0 = 0 and the slope is cosh⁡0=1\cosh 0 = 1, so the tangent line there is y=xy = x.

Hyperbolic cosine

Definition The function cosh⁡\cosh is even, defined and differentiable on R\mathbb{R}, with range [1,∞)[1, \infty), and

ddxcosh⁡x=ex−e−x2=sinh⁡x.\frac{d}{dx} \cosh x = \frac{e^x - e^{-x}}{2} = \sinh x.

In plain words Unlike cosine, there is no minus sign here. Since sinh⁡x\sinh x is negative for x<0x < 0 and positive for x>0x > 0, the graph of cosh⁡\cosh falls, bottoms out at height 1 when x=0x = 0, then rises.

Example By the chain rule

ddxcosh⁡(3x)=3sinh⁡(3x).\frac{d}{dx} \cosh(3x) = 3 \sinh(3x).

At x=ln⁡2x = \ln 2 the slope of cosh⁡x\cosh x itself is sinh⁡(ln⁡2)=34\sinh(\ln 2) = \frac{3}{4}.

Identities

Definition For every real xx,

cosh⁡2x−sinh⁡2x=1,sinh⁡(2x)=2sinh⁡xcosh⁡x,cosh⁡(2x)=cosh⁡2x+sinh⁡2x,\cosh^2 x - \sinh^2 x = 1, \qquad \sinh(2x) = 2 \sinh x \cosh x, \qquad \cosh(2x) = \cosh^2 x + \sinh^2 x,

and, dividing the first by cosh⁡2x\cosh^2 x,

1−tanh⁡2x=sech⁡2x.1 - \tanh^2 x = \operatorname{sech}^2 x.

Each follows by expanding the exponential definitions.

In plain words The point (cosh⁡t,sinh⁡t)(\cosh t, \sinh t) lies on the hyperbola x2−y2=1x^2 - y^2 = 1, just as (cos⁡t,sin⁡t)(\cos t, \sin t) lies on the circle x2+y2=1x^2 + y^2 = 1. That is where the name comes from.

Example At x=ln⁡2x = \ln 2 the first identity reads

cosh⁡2x−sinh⁡2x=2516−916=1.\cosh^2 x - \sinh^2 x = \frac{25}{16} - \frac{9}{16} = 1.

For the second, 2x=ln⁡42x = \ln 4 and

sinh⁡(ln⁡4)=4−1/42=158=2⋅34⋅54.\sinh(\ln 4) = \frac{4 - 1/4}{2} = \frac{15}{8} = 2 \cdot \frac{3}{4} \cdot \frac{5}{4}.

Hyperbolic tangent

Definition For every real xx the hyperbolic tangent is

tanh⁡x=sinh⁡xcosh⁡x=ex−e−xex+e−x.\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}.

It is odd, differentiable on R\mathbb{R}, has range (−1,1)(-1, 1), and

ddxtanh⁡x=cosh⁡2x−sinh⁡2xcosh⁡2x=sech⁡2x.\frac{d}{dx} \tanh x = \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x} = \operatorname{sech}^2 x.

In plain words The graph of tanh⁡\tanh is an S shape that rises from −1-1 to 11 and has horizontal asymptotes y=±1y = \pm 1.

Example At ln⁡2\ln 2 the value is tanh⁡(ln⁡2)=3/45/4=35\tanh(\ln 2) = \frac{3/4}{5/4} = \frac{3}{5}, and the slope there is

1−tanh⁡2(ln⁡2)=1−925=1625.1 - \tanh^2(\ln 2) = 1 - \frac{9}{25} = \frac{16}{25}.

Reciprocal hyperbolic functions

Definition The other three hyperbolic functions and their derivatives are

sech⁡x=1cosh⁡x,x∈R,(sech⁡x)′=−sech⁡xtanh⁡x,\operatorname{sech} x = \frac{1}{\cosh x}, \quad x \in \mathbb{R}, \qquad (\operatorname{sech} x)' = -\operatorname{sech} x \tanh x,csch⁡x=1sinh⁡x,x≠0,(csch⁡x)′=−csch⁡xcoth⁡x,\operatorname{csch} x = \frac{1}{\sinh x}, \quad x \neq 0, \qquad (\operatorname{csch} x)' = -\operatorname{csch} x \coth x,coth⁡x=cosh⁡xsinh⁡x,x≠0,(coth⁡x)′=−csch⁡2x.\coth x = \frac{\cosh x}{\sinh x}, \quad x \neq 0, \qquad (\coth x)' = -\operatorname{csch}^2 x.

In plain words These follow from the quotient rule. Notice that all three carry a minus sign, while among the trigonometric functions sec⁡\sec had none.

Example At x=ln⁡2x = \ln 2 we have coth⁡x=53\coth x = \frac{5}{3} and csch⁡x=43\operatorname{csch} x = \frac{4}{3}, so (coth⁡x)′=−169(\coth x)' = -\frac{16}{9}.

Catenary

Definition A catenary is a curve of the form

y=acosh⁡(xa),a>0,y = a \cosh\left(\frac{x}{a}\right), \qquad a > 0,

up to shifting. It is the shape of a uniform, flexible chain hanging under its own weight. Its slope is y′=sinh⁡(xa)y' = \sinh\left(\frac{x}{a}\right), and it satisfies the differential equation

y′′=1a1+(y′)2.y'' = \frac{1}{a} \sqrt{1 + (y')^2}.

In plain words The parameter aa sets how tight the chain is. Large aa gives a flat, shallow sag and small aa a deep one.

Example For a=2a = 2 the curve is y=2cosh⁡(x2)y = 2 \cosh\left(\frac{x}{2}\right). At x=2x = 2 the height is 2cosh⁡1≈3.0862 \cosh 1 \approx 3.086 and the slope is sinh⁡1≈1.175\sinh 1 \approx 1.175. The second derivative there is 12cosh⁡1\frac{1}{2} \cosh 1, which matches 121+sinh⁡21\frac{1}{2}\sqrt{1 + \sinh^2 1} by the identity cosh⁡2−sinh⁡2=1\cosh^2 - \sinh^2 = 1.

Inverse hyperbolic functions

Definition The inverse hyperbolic functions have logarithm formulas,

arsinh⁡x=ln⁡(x+x2+1),x∈R,\operatorname{arsinh} x = \ln\left(x + \sqrt{x^2 + 1}\right), \quad x \in \mathbb{R},arcosh⁡x=ln⁡(x+x2−1),x≥1,\operatorname{arcosh} x = \ln\left(x + \sqrt{x^2 - 1}\right), \quad x \ge 1,artanh⁡x=12ln⁡(1+x1−x),−1<x<1.\operatorname{artanh} x = \frac{1}{2} \ln\left(\frac{1 + x}{1 - x}\right), \quad -1 < x < 1.

Since cosh⁡\cosh is not one-to-one, arcosh⁡\operatorname{arcosh} is the inverse of cosh⁡\cosh restricted to [0,∞)[0, \infty), so its values are in [0,∞)[0, \infty). The reciprocal functions have inverses too, arcoth⁡x=artanh⁡1x\operatorname{arcoth} x = \operatorname{artanh}\frac{1}{x} for ∣x∣>1\lvert x \rvert > 1, arsech⁡x=arcosh⁡1x\operatorname{arsech} x = \operatorname{arcosh}\frac{1}{x} for 0<x≤10 < x \le 1, and arcsch⁡x=arsinh⁡1x\operatorname{arcsch} x = \operatorname{arsinh}\frac{1}{x} for x≠0x \neq 0. These are also written sinh⁡−1x\sinh^{-1} x, cosh⁡−1x\cosh^{-1} x and tanh⁡−1x\tanh^{-1} x. The formula for arsinh⁡\operatorname{arsinh} comes from solving

x=ey−e−y2,x = \frac{e^y - e^{-y}}{2},

which becomes the quadratic (ey)2−2xey−1=0(e^y)^2 - 2x e^y - 1 = 0 in eye^y, and keeping the positive root ey=x+x2+1e^y = x + \sqrt{x^2 + 1}.

In plain words Because the hyperbolic functions are built from exe^x, their inverses are built from ln⁡\ln.

Example Since (3/4)2+1=54\sqrt{(3/4)^2 + 1} = \frac{5}{4}, the formula gives

arsinh⁡(34)=ln⁡(34+54)=ln⁡2,\operatorname{arsinh}\left(\frac{3}{4}\right) = \ln\left(\frac{3}{4} + \frac{5}{4}\right) = \ln 2,

which matches sinh⁡(ln⁡2)=34\sinh(\ln 2) = \frac{3}{4}. Similarly (5/4)2−1=34\sqrt{(5/4)^2 - 1} = \frac{3}{4}, so

arcosh⁡(54)=ln⁡(54+34)=ln⁡2.\operatorname{arcosh}\left(\frac{5}{4}\right) = \ln\left(\frac{5}{4} + \frac{3}{4}\right) = \ln 2.

Derivative formulas and real domains

Definition Each inverse is differentiable on the open set shown.

FunctionDerivativeValid for
arsinh⁡x\operatorname{arsinh} x1x2+1\frac{1}{\sqrt{x^2 + 1}}all real xx
arcosh⁡x\operatorname{arcosh} x1x2−1\frac{1}{\sqrt{x^2 - 1}}x>1x > 1
artanh⁡x\operatorname{artanh} x11−x2\frac{1}{1 - x^2}∣x∣<1\lvert x \rvert < 1
arcoth⁡x\operatorname{arcoth} x11−x2\frac{1}{1 - x^2}∣x∣>1\lvert x \rvert > 1
arsech⁡x\operatorname{arsech} x−1x1−x2-\frac{1}{x\sqrt{1 - x^2}}0<x<10 < x < 1
arcsch⁡x\operatorname{arcsch} x−1∣x∣1+x2-\frac{1}{\lvert x \rvert \sqrt{1 + x^2}}x≠0x \neq 0

The function arcosh⁡\operatorname{arcosh} is defined at x=1x = 1 but not differentiable there, because sinh⁡0=0\sinh 0 = 0 and the graph has a vertical tangent.

In plain words Each formula comes from the inverse function rule plus an identity. The domain column matters, since artanh⁡\operatorname{artanh} and arcoth⁡\operatorname{arcoth} share a formula but live on different sets.

Example By the table, the slope of arsinh⁡\operatorname{arsinh} at 34\frac{3}{4} is

19/16+1=15/4=45.\frac{1}{\sqrt{9/16 + 1}} = \frac{1}{5/4} = \frac{4}{5}.

The inverse function rule gives the same, since arsinh⁡34=ln⁡2\operatorname{arsinh}\frac{3}{4} = \ln 2 and 1/cosh⁡(ln⁡2)=451 / \cosh(\ln 2) = \frac{4}{5}.

Check yourself

  1. Differentiate y=sinh⁡(x2)y = \sinh(x^2).
Show the answer

By the chain rule y′=2xcosh⁡(x2)y' = 2x \cosh(x^2).

  1. Show that ddxln⁡(cosh⁡x)=tanh⁡x\frac{d}{dx} \ln(\cosh x) = \tanh x and say where this holds.
Show the answer

By the chain rule

ddxln⁡(cosh⁡x)=sinh⁡xcosh⁡x=tanh⁡x.\frac{d}{dx} \ln(\cosh x) = \frac{\sinh x}{\cosh x} = \tanh x.

This holds for every real xx, since cosh⁡x≥1>0\cosh x \ge 1 > 0.

  1. Find the slope of y=artanh⁡xy = \operatorname{artanh} x at x=12x = \frac{1}{2}.
Show the answer

The derivative of artanh⁡x\operatorname{artanh} x is 11−x2\frac{1}{1 - x^2} for ∣x∣<1\lvert x \rvert < 1, so the slope is

11−1/4=43.\frac{1}{1 - 1/4} = \frac{4}{3}.

How this lesson connects

It also uses hyperbolic functions.