\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Derivatives - Inverse Hyperbolic Functions

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On this page we discuss the derivative of inverse hyperbolic functions. We discussed the basics of hyperbolic functions on the hyperbolic derivatives page.

Since you know from the discussion on the hyperbolics page that the hyperbolic trig functions are based on the exponential function \(e^x\), you would be correct if you thought that the inverse hyperbolic trig functions are based on the natural logarithm function \(\ln(x)\). Let's look at each function individually and work some practice problems as we go.

Inverse Hyperbolic Sine - \( y = \arcsinh(x) \)

From the hyperbolic functions page, the equation for hyperbolic sine is \[\displaystyle{ y = \sinh(x) = \frac{e^x-e^{-x}}{2} }\] We can use this definition to get the equation for inverse hyperbolic sine in terms of natural logarithms. Try this on your own before looking at the solution.

Use the definition of \( y = \sinh(x) \) to derive the expression for \( y = \text{arcsinh}(x) \).

Problem Statement

Use the definition of \( y = \sinh(x) \) to derive the expression for \( y = \text{arcsinh}(x) \).

Solution

2319 video

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So you now know that the inverse hyperbolic sine function can be written as \(\displaystyle{ \arcsinh(x) = \ln(x + \sqrt{x^2+1}) }\). Now work this next practice problem to calculate the derivative of the inverse hyperbolic sine function using this function.

Use the definition of \( y = \text{arcsinh}(x) \) to calculate \( dy/dx \).

Problem Statement

Use the definition of \( y = \text{arcsinh}(x) \) to calculate \( dy/dx \).

Solution

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Okay, in summary, here are the definition of the inverse hyperbolic sine function and it's derivative.

Inverse Hyperbolic Sine Function and Derivative

\(\displaystyle{ \arcsinh(x) = \ln(x + \sqrt{x^2+1}) }\)

\(\displaystyle{ \frac{d}{dx}[ \arcsinh(x)] = \frac{1}{\sqrt{x^2+1}} }\)

As is common with inverse functions, sometimes you will see the notation \(\sinh^{-1}(x)\) instead of \( \arcsinh(x)\). Remember that \(\displaystyle{ \sinh^{-1}(x) \neq \frac{1}{\sinh(x)} }\). This notation may be used by your instructor, in your textbook and in some of the practice problems on this site.

Here is a video clip proving the derivative of \(\sinh^{-1}(x)\), which shows you how to manipulate inverse hyperbolic functions.

Krista King Math - Inverse Hyperbolic Functions [5min-10secs]

video by Krista King Math

Figure 1 - \( y = \arcsinh(x) \)

built with GeoGebra

The plot of the inverse hyperbolic sine function (thicker green line) is shown in figure 1. The thin red line in the plot is \( y = \sinh(x) \). Since \( y = \arcsinh(x) \) is the inverse function of \( y = \sinh(x) \), it is of course a reflection about the line \( y = x \), also shown in this plot. Notice that the inverse hyperbolic sine function is defined for all x, i.e. the domain is \( -\infty < x < \infty \). The range is also the set of all real numbers.

Now let's work additional practice problems involving inverse hyperbolic sine. Unless otherwise instructed, calculate the derivative of these functions.

Calculate the derivative of \(\displaystyle{y=x^2\sinh^{-1}(2x)}\)

Problem Statement

Calculate the derivative of \(\displaystyle{y=x^2\sinh^{-1}(2x)}\)

Solution

1095 video

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Calculate the derivative of \(\displaystyle{f(x)=(\sinh^{-1}(x))^{3/2}}\)

Problem Statement

Calculate the derivative of \(\displaystyle{f(x)=(\sinh^{-1}(x))^{3/2}}\)

Solution

1100 video

video by Krista King Math

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Calculate the derivative of \(\displaystyle{y=x\sinh^{-1}(x/3)-\sqrt{9+x^2}}\)

Problem Statement

Calculate the derivative of \(\displaystyle{y=x\sinh^{-1}(x/3)-\sqrt{9+x^2}}\)

Solution

1101 video

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Inverse Hyperbolic Cosine - \( y = \arccosh(x) \)

Now let's do the same thing with inverse hyperbolic cosine.

Use the definition of \( y = \cosh(x) \) to derive the expression for \( y = \text{arccosh}(x) \).

Problem Statement

Use the definition of \( y = \cosh(x) \) to derive the expression for \( y = \text{arccosh}(x) \).

Solution

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Use the definition of \( y = \text{arccosh}(x) \) to calculate \( dy/dx \).

Problem Statement

Use the definition of \( y = \text{arccosh}(x) \) to calculate \( dy/dx \).

Solution

2322 video

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In summary, here are the definition of the inverse hyperbolic cosine function and it's derivative.

Inverse Hyperbolic Cosine Function and Derivative

\(\displaystyle{ \arccosh(x) = \ln(x + \sqrt{x^2-1}) }\)

\(\displaystyle{ \frac{d}{dx}[ \arccosh(x)] = \frac{1}{\sqrt{x^2-1}} }\)

Figure 2 - \( y = \arccosh(x) \)

built with GeoGebra

Figure 2 shows the inverse hyperbolic cosine function (thicker green line), the hyperbolic cosine function (thin red line) and \(y=x\). Since \(y=\cosh(x)\) is not one-to-one (does not pass the horizontal line test), we need to restrict the domain to be able to define an inverse function. So the domain of \( y = \arccosh(x) \) is \( x \geq 1 \). The range is \( y \geq 0 \).

Before we go on, let's compare the first two inverse hyperbolic functions.

Figure 3 - Comparing \(\arcsinh(x)\) and \(\arccosh(x)\)

built with GeoGebra

\(\displaystyle{ \arcsinh(x) = \ln(x + \sqrt{x^2+1}) }\)

\(\displaystyle{ \arccosh(x) = \ln(x + \sqrt{x^2-1}) }\)

The equations for these two functions are very similar, differing only by a minus sign, so it is interesting to compare them. Notice from Figure 3, that as x increases the plots seem to approach one another. This can be confirmed by the equations. Notice that as x increases it is squared in the square root term and the constant 1 becomes negligible when compared to it. So both plots approach \(y=\ln(2x)\), which is also shown in Figure 3 (red line).

An interesting observation is that the two functions are approaching one another very quickly. Figure 3 is plotted to only about \(x=5\) and already we can barely tell the difference between the three plots.

Now let's work additional practice problems involving inverse hyperbolic cosine. Unless otherwise instructed, calculate the derivative of these functions.

Calculate the derivative of \(f(x) = 2\cosh^{-1}(5x) \).

Problem Statement

Calculate the derivative of \(f(x) = 2\cosh^{-1}(5x) \).

Solution

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So if you are thinking that since the inverse hyperbolic sine and cosine are so similar, the other inverse hyperbolic functions also come in similar pairs, you would be correct. So we will now cover the remaining functions in pairs.

Inverse Hyperbolic Tangent and Cotangent

First, let's derive the equations for the functions themselves.

Use definitions of \( \sinh(x) \) and \( \cosh(x) \) to derive the equation for \( \text{arctanh}(x) \).

Problem Statement

Use definitions of \( \sinh(x) \) and \( \cosh(x) \) to derive the equation for \( \text{arctanh}(x) \).

Hint

\(\displaystyle{ \sinh(x) = \frac{e^x-e^{-x}}{2} }\)
\(\displaystyle{ \cosh(x) = \frac{e^x+e^{-x}}{2} }\)
Set up an expression for \( \tanh(x) \) and work from there.

Problem Statement

Use definitions of \( \sinh(x) \) and \( \cosh(x) \) to derive the equation for \( \text{arctanh}(x) \).

Hint

\(\displaystyle{ \sinh(x) = \frac{e^x-e^{-x}}{2} }\)
\(\displaystyle{ \cosh(x) = \frac{e^x+e^{-x}}{2} }\)
Set up an expression for \( \tanh(x) \) and work from there.

Solution

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Use the definition of \( \text{arctanh}(x) \) to derive the equation for \( \text{arccoth}(x) \).

Problem Statement

Use the definition of \( \text{arctanh}(x) \) to derive the equation for \( \text{arccoth}(x) \).

Hint

\( \text{arccoth}(x) = \text{arctanh}(1/x) \)

Problem Statement

Use the definition of \( \text{arctanh}(x) \) to derive the equation for \( \text{arccoth}(x) \).

Hint

\( \text{arccoth}(x) = \text{arctanh}(1/x) \)

Solution

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Figure 4 - \( y = \arctanh(x) \) [green], \( y = \arccoth(x) \) [orange]

built with GeoGebra

Let's look at the equations and plots in detail, shown in Figure 4.
First, notice that the domains of both functions are not all real numbers.
For inverse hyperbolic tangent the equation is \[ \displaystyle{ \arctanh(x) = \frac{1}{2}\ln\left[ \frac{1+x}{1-x} \right] } \] The natural logarithm function requires that \(\displaystyle{ \frac{1+x}{1-x} }\) must be positive. From the equation, it looks like \( x = \pm 1 \) are important points. So let's look at three interval, \( x < -1 \), \( -1 < x < 1 \) and \( x > 1 \).
Note - We can reject \( x=1 \) and \( x=-1\) immediately. [why?]

Let's look at the first interval, \( x < -1 \). In this case, the denominator of \(\displaystyle{ \frac{1+x}{1-x} }\) is positive but the numerator is negative. This means that the fraction is negative, which is not allowed by the natural logarithm function. So \( x < -1 \) is not part of the domain.
The interval \( -1 < x < 1 \) looks okay since the numerator and denominator are both positive for all values in this interval.
Finally, the interval \( x > 1 \) causes problems since the numerator is positive but the denominator is negative, thus the fraction is negative. Again, not allowed for the natural logarithm. (To check these conclusions, plot \(\displaystyle{ y=\frac{1+x}{1-x} }\) and determine the x-values where the \( y > 0 \).)
So our conclusion is that the domain of \(y = \arctanh(x)\) is \( -1 < x < 1 \), which can be written \( |x| < 1 \).

We can use similar logic to get the domain of \( y = \arccoth(x) \). However, let's take a shortcut. Notice that the fraction inside the inverse hyperbolic cotangent function is just the negative of the fraction inside the inverse hyperbolic tangent function. Let's compare the fractions.

Inside Fraction Comparison

inverse hyperbolic tangent

\(\displaystyle{ y = \frac{1+x}{1-x} }\)

inverse hyperbolic cotangent

\(\displaystyle{ y = \frac{x+1}{x-1} }\)

The numerators are the same, the denominators are negatives of each other, which means that the fractions are negatives of each other. So the opposite conclusions can be made about the domain. This means that the domain of \( y = \arccoth(x) \) is \(|x|>1\).

Now let's calculate the derivatives.

Calculate the derivative of \( \text{arctanh}(x) \) from the definition.

Problem Statement

Calculate the derivative of \( \text{arctanh}(x) \) from the definition.

Solution

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Calculate the derivative of \( \text{arccoth}(x) \) from the definition.

Problem Statement

Calculate the derivative of \( \text{arccoth}(x) \) from the definition.

Solution

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Notice that the equations for the derivatives look the same but, the domains transfer directly from each function, which means that the derivatives are different. Here is a summary of the results.

Inverse Hyperbolic Tangent and Cotangent Functions and Derivatives

\(\displaystyle{ \arctanh(x) = \frac{1}{2}\ln\left[ \frac{1+x}{1-x} \right] }\) for \( |x| < 1 \)

\(\displaystyle{ \frac{d}{dx}[ \arctanh(x)] = \frac{1}{(1-x^2)} }\) for \( |x| < 1 \)

\(\displaystyle{ \arccoth(x) = \frac{1}{2}\ln\left[ \frac{x+1}{x-1} \right] }\) for \( x > 1 \)

\(\displaystyle{ \frac{d}{dx}[ \arccoth(x)] = \frac{1}{1-x^2} }\) for \( x > 1 \)

Let's work some practice problems containing the inverse hyperbolic tangent and cotangent.

Calculate the derivative of \( y = \tanh^{-1}\sqrt{x} \).

Problem Statement

Calculate the derivative of \( y = \tanh^{-1}\sqrt{x} \).

Solution

1096 video

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Calculate the derivative of \( \text{arctanh}(4x) \).

Problem Statement

Calculate the derivative of \( \text{arctanh}(4x) \).

Solution

2323 video

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Inverse Hyperbolic Secant and Cosecant Functions

The inverse hyperbolic secant and cosecant can be analysed in a similar manner as we did for tangent and cotangent. So we will allow you to work through the practice problems and show the results.

Use the definition of \( \text{arccosh}(x) \) to derive the equation for \( \text{arcsech}(x) \).

Problem Statement

Use the definition of \( \text{arccosh}(x) \) to derive the equation for \( \text{arcsech}(x) \).

Hint

\( \text{arcsech}(x) = \text{arccosh}(1/x) \)

Problem Statement

Use the definition of \( \text{arccosh}(x) \) to derive the equation for \( \text{arcsech}(x) \).

Hint

\( \text{arcsech}(x) = \text{arccosh}(1/x) \)

Solution

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Use the definition of \( \text{arcsinh}(x) \) to derive the equation for \( \text{arccsch}(x) \).

Problem Statement

Use the definition of \( \text{arcsinh}(x) \) to derive the equation for \( \text{arccsch}(x) \).

Hint

\( \text{arccsch}(x) = \text{arcsinh}(1/x) \)

Problem Statement

Use the definition of \( \text{arcsinh}(x) \) to derive the equation for \( \text{arccsch}(x) \).

Hint

\( \text{arccsch}(x) = \text{arcsinh}(1/x) \)

Solution

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Use the definition of \( y = \text{arcsech}(x) \) to calculate \( dy/dx \).

Problem Statement

Use the definition of \( y = \text{arcsech}(x) \) to calculate \( dy/dx \).

Solution

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Use the definition of \( y = \text{arccsch}(x) \) to calculate \(dy/dx\).

Problem Statement

Use the definition of \( y = \text{arccsch}(x) \) to calculate \(dy/dx\).

Solution

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Inverse Hyperbolic Secant Derivative

Inverse Hyperbolic Secant Function and Derivative

\(\displaystyle{ \arcsech(x) = \ln \left[ \frac{1}{x}(1+\sqrt{1-x^2}) \right] }\) for \( 0 < x \leq 1 \)

\(\displaystyle{ \frac{d}{dx}[ \arcsech(x)] = \frac{-1}{x\sqrt{1-x^2}} }\) for \( 0 < x \leq 1 \)

Calculate the derivative of \(\displaystyle{y=\text{sech}^{-1}\sqrt{1-x^2}}\)

Problem Statement

Calculate the derivative of \(\displaystyle{y=\text{sech}^{-1}\sqrt{1-x^2}}\)

Solution

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Inverse Hyperbolic Cosecant Derivative

Inverse Hyperbolic Cosecant Function and Derivative

\(\displaystyle{ \arccsch(x) = \ln \left[ \frac{1}{x}(1+\sqrt{1+x^2}) \right] }\) for \( x \neq 0 \)

\(\displaystyle{ \frac{d}{dx}[ \arccsch(x)] = \frac{-1}{|x|\sqrt{1+x^2}} }\) for \( x \neq 0 \)

Calculate the derivative of \( (\text{csch}^{-1}(x))^2 \).

Problem Statement

Calculate the derivative of \( (\text{csch}^{-1}(x))^2 \).

Solution

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You CAN Ace Calculus

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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