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# Calculus

Differentiation of Functions

# Derivatives of Hyperbolic Functions

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Problems 1-3
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Problems 4-20

### Derivatives of Hyperbolic Functions

The derivatives of hyperbolic functions can be easily found as these functions are combinations of $${e^x}$$ and $${e^{ – x}}$$. For example, the hyperbolic sine and hyperbolic cosine functions are defined by

${\sinh x = \frac{{{e^x} – {e^{ – x}}}}{2},\;\;\;}\kern-0.3pt{\cosh x = \frac{{{e^x} + {e^{ – x}}}}{2}.}$

The derivative of these functions are

${{\left( {\sinh x} \right)^\prime } = {\left( {\frac{{{e^x} – {e^{ – x}}}}{2}} \right)^\prime } } = {\frac{{{e^x} + {e^{ – x}}}}{2} = \cosh x,\;\;\;}\kern-0.3pt {{\left( {\cosh x} \right)^\prime } = {\left( {\frac{{{e^x} + {e^{ – x}}}}{2}} \right)^\prime } } = {\frac{{{e^x} – {e^{ – x}}}}{2} = \sinh x.}$

We derive the derivative formula for the hyperbolic tangent:

${{\left( {\tanh x} \right)^\prime } = {\left( {\frac{{\sinh x}}{{\cosh x}}} \right)^\prime } } = {\frac{{{{\left( {\sinh x} \right)}^\prime }\cosh x – \sinh x{{\left( {\cosh x} \right)}^\prime }}}{{{{\cosh }^2}x}} } = {\frac{{\cosh x \cdot \cosh x – \sinh x \cdot \sinh x}}{{{{\cosh }^2}x}} } = {\frac{{{{\cosh }^2}x – {{\sinh }^2}x}}{{{{\cosh }^2}x}}.}$

It is known that the hyperbolic sine and cosine are connected by the relationship

${\cosh ^2}x – {\sinh ^2}x = 1.$

Therefore, the derivative of the hyperbolic tangent is written as

${{\left( {\tanh x} \right)^\prime } }={ \frac{{{{\cosh }^2}x – {{\sinh }^2}x}}{{{{\cosh }^2}x}} } = {\frac{1}{{{{\cosh }^2}x}} } = {{\text{sech}^2}x.}$

Similarly, we can obtain the differentiation formulas for the other hyperbolic functions:

${{\left( {\coth x} \right)^\prime } }={ {\left( {\frac{{\cosh x}}{{\sinh x}}} \right)^\prime } } = {\frac{{{{\left( {\cosh x} \right)}^\prime }\sinh x – \cosh x{{\left( {\sinh x} \right)}^\prime }}}{{{{\sinh }^2}x}} } = { – \frac{{{{\cosh }^2}x – {{\sinh }^2}x}}{{{{\sinh }^2}x}} } = { – \frac{1}{{{{\sinh }^2}x}} } = { – {\text{csch}^2}x,}$
${{\left( {\text{sech}\,x} \right)^\prime } }={ {\left( {\frac{1}{{\cosh x}}} \right)^\prime } } = { – \frac{1}{{{{\cosh }^2}x}} \cdot {\left( {\cosh x} \right)^\prime } } = { – \frac{1}{{{{\cosh }^2}x}} \cdot \sinh x } = { – \frac{1}{{\cosh x}} \cdot \frac{{\sinh x}}{{\cosh x}} } = { – \text{sech}\,x\tanh x,}$
${{\left( {\text{csch}\,x} \right)^\prime } }={ {\left( {\frac{1}{{\sinh x}}} \right)^\prime } } = { – \frac{1}{{{\sinh^2}x}} \cdot {\left( {\sinh x} \right)^\prime } } = { – \frac{1}{{{\sinh^2}x}} \cdot \cosh x } = { – \frac{1}{{\sinh x}} \cdot \frac{{\cosh x}}{{\sinh x}} } = { – \text{csch}\,x\coth x\;\;}\kern-0.3pt{\left( {x \ne 0} \right).}$

As can be seen, the derivatives of the hyperbolic functions are very similar to the derivatives of trigonometric functions. However, it is important to note the difference in signs! If the derivative of the cosine function is given by

${\left( {\cos x} \right)^\prime } = – \sin x,$

then the minus sign is missing for the derivative of the hyperbolic cosine:

${\left( {\cosh x} \right)^\prime } = \sinh x.$

For the secant function, the situation with the sign is exactly reversed:

${{{\left( {\sec x} \right)^\prime } = \sec x\tan x,\;\;\;}}\kern-0.3pt {{{\left( {\text{sech}\,x} \right)^\prime } }={ – \text{sech}\,x\tanh x.}}$

### Derivatives of Inverse Hyperbolic Functions

Consider now the derivatives of $$6$$ inverse hyperbolic functions. The corresponding differentiation formulas can be derived using the inverse function theorem.

Take, for example, the function $$y = f\left( x \right)$$ $$= \text{arcsinh}\,x$$ (inverse hyperbolic sine). Together with the function $$x = \varphi \left( y \right)$$ $$= \sinh y$$ they form a pair of mutually inverse funtions. Then the derivative of the inverse hyperbolic sine is given by

${{\left( {\text{arcsinh}\,x} \right)^\prime } }={ f’\left( x \right) } = {\frac{1}{{\varphi’\left( y \right)}} } = {\frac{1}{{{{\left( {\sinh y} \right)}^\prime }}} } = {\frac{1}{{\cosh y}} } = {\frac{1}{{\sqrt {1 + {\sinh^2}y} }} } = {\frac{1}{{\sqrt {1 + {\sinh^2}\left( {\text{arcsinh}\,x} \right)} }} } = {\frac{1}{{\sqrt {1 + {x^2}} }}.}$

Similarly, we can derive the derivatives for the inverse hyperbolic cosine, tangent and cotangent functions.

${{\left( {\text{arccosh}\,x} \right)^\prime } }={ f’\left( x \right) } = {\frac{1}{{\varphi’\left( y \right)}} } = {\frac{1}{{{{\left( {\cosh y} \right)}^\prime }}} } = {\frac{1}{{\sinh y}} } = {\frac{1}{{\sqrt {{\cosh^2}y – 1} }} } = {\frac{1}{{\sqrt {{\cosh^2}\left( {\text{arccosh}\,x} \right) – 1} }} } = {\frac{1}{{\sqrt {{x^2} – 1} }}\;\;\left( {x \gt 1} \right),}$
${{\left( {\text{arctanh}\,x} \right)^\prime } }={ f’\left( x \right) } = {\frac{1}{{\varphi’\left( y \right)}} } = {\frac{1}{{{{\left( {\tanh y} \right)}^\prime }}} } = {\frac{1}{{\frac{1}{{{{\cosh }^2}y}}}} } = {{\cosh ^2}y.}$

The identity $${\cosh ^2}y – {\sinh ^2}y = 1$$ implies that

${1 – {\tanh ^2}y = \frac{1}{{{{\cosh }^2}y}}\;\;}\kern-0.3pt {\text{or}\;\;{\cosh ^2}y = \frac{1}{{1 – {{\tanh }^2}y}}.}$

Therefore

${{\left( {\text{arctanh}\,x} \right)^\prime } }={ {\cosh ^2}y }={ \frac{1}{{1 – {{\tanh }^2}y}} } = {\frac{1}{{1 – {{\tanh }^2}\left( {\text{arctanh}\,x} \right)}} } = {\frac{1}{{1 – {x^2}}}\;\;\left( {\left| x \right| \lt 1} \right).}$

In a similar way, we find the derivative of the function $$y = f\left( x \right) = \text{arccoth}\,x$$ (inverse hyperbolic cotangent):

${{\left( {\text{arccoth}\,x} \right)^\prime } }={ f’\left( x \right) } = {\frac{1}{{\varphi’\left( y \right)}} } = {\frac{1}{{{{\left( {\coth y} \right)}^\prime }}} } = {\frac{1}{{\left( { – \frac{1}{{{{\sinh }^2}y}}} \right)}} } = { – {\sinh ^2}y.}$

Given that

${{\coth ^2}y – 1 = \frac{1}{{{{\sinh }^2}y}},\;\;}\Rightarrow {{\sinh ^2}y = \frac{1}{{{{\coth }^2}y – 1}},}$

we obtain

${\left( {\text{arccoth}\,x} \right)^\prime } ={ – {\sinh ^2}y } = { – \frac{1}{{{{\coth }^2}y – 1}} } = { – \frac{1}{{{{\coth }^2}\left( {\text{arccoth}\,x} \right) – 1}} } = { – \frac{1}{{{x^2} – 1}} } = {\frac{1}{{1 – {x^2}}}\;\;}\kern-0.3pt{\left( {\left| x \right| \gt 1} \right).}$

As can be seen, the derivatives of the functions $$\text{arctanh}\,x$$ and $$\text{arccoth}\,x$$ are the same, but they are determined for different values of $$x.$$ The domain restrictions for the inverse hyperbolic tangent and cotangent follow from the range of the functions $$y = \tanh x$$ and $$y = \coth x,$$ respectively.

We also derive the derivatives of the inverse hyperbolic secant and cosecant, though these functions are rare.

In accordance with the described algorithm, we write two mutually inverse functions: $$y = f\left( x \right) = \text{arcsech}\,x$$ $$\left( {x \in \left( {0,1} \right]} \right)$$ and $$x = \varphi \left( y \right) = \text{sech}\,y$$ $$\left( {y \gt 0} \right)$$.

Calculate the derivative:

${{\left( {\text{arcsech}\,x} \right)^\prime } }={ f’\left( x \right) } = {\frac{1}{{\varphi’\left( y \right)}} } = {\frac{1}{{{{\left( {\text{sech}\,y} \right)}^\prime }}} } = {\frac{1}{{\text{sech}\,y\tanh y}}.}$

Express $$\tanh y$$ in terms of $$\text{sech}\,y$$ given that $$y \gt 0$$:

${{\cosh ^2}y – {\sinh ^2}y = 1,\;\;}\Rightarrow {1 – {\tanh ^2}y = \frac{1}{{{{\cosh }^2}y}} = {\text{sech}^2}y,\;\;}\Rightarrow {{\tanh ^2}y = 1 – {\text{sech}^2}y,\;\;}\Rightarrow {\tanh y = \sqrt {1 – {{\text{sech}}^2}y}.}$

Then the result is

${{\left( {\text{arcsech}\,x} \right)^\prime } }={ – \frac{1}{{\text{sech}\,y \tanh y}} } = { – \frac{1}{{x\sqrt {1 – {x^2}} }},\;\;}\kern-0.3pt{x \in \left( {0,1} \right).}$

Similarly, we can find the derivative of the inverse hyperbolic cosecant. Suppose that $$y = f\left( x \right)$$ $$= \text{arccsch}\,x$$ $$\left( {x \in \mathbb{R},\;x \ne 0} \right)$$ and $$x = \varphi \left( y \right)$$ $$= \text{csch}\,y$$ $$\left( {y \ne 0} \right)$$.

We first consider the branch $$x \gt 0$$. In this case, the variable $$y$$ takes the values $$y \gt 0$$ (the graphs of these functions can be found on the Functions and Their Graphs page). The derivative of the inverse hyperbolic cosecant is expressed as

${{\left( {\text{arccsch}\,x} \right)^\prime } }={ f’\left( x \right) } = {\frac{1}{{\varphi’\left( y \right)}} } = {\frac{1}{{{{\left( {\text{csch}\,y} \right)}^\prime }}} } = { – \frac{1}{{\text{csch}\,y\coth y}}.}$

Make the substitution

${{\cosh ^2}y – {\sinh ^2}y = 1,\;\;}\Rightarrow {{\coth ^2}y – 1 = \frac{1}{{{{\sinh }^2}y}} = {\text{csch}^2}y,\;\;}\Rightarrow {{\coth ^2}y = 1 + {\text{csch}^2}y,\;\;}\Rightarrow {\coth y = \pm \sqrt {1 + {{\text{csch}}^2}y}.}$

Given that $$y \gt 0,$$ we choose the “+” sign in front of the square root. Consequently,

${{\left( {\text{arccsch}\,x} \right)^\prime } }={ – \frac{1}{{\text{csch}\,y \coth y}} } = { – \frac{1}{{x\sqrt {1 + {x^2}} }}\;\;}\kern-0.3pt{\left( {x \gt 0} \right).}$

Now we consider a pair of mutually inverse functions for $$x \lt 0$$. Due to the oddness of the hyperbolic cosecant, this corresponds to the condition $$y \lt 0$$. Moreover, the hyperbolic cosecant is also negative for $$y \lt 0$$: $$\coth y \gt 0$$, i.e. in this case it is necessary to write the hyperbolic identity as

${\coth y = – \sqrt {1 + {{\text{csch}}^2}y} \;\;}\kern-0.3pt{\left( {y \lt 0} \right).}$

Then the derivative of the inverse hyperbolic cosecant for $$x \lt 0$$ is given by

${{\left( {\text{arccsch}\,x} \right)^\prime } }={ – \frac{1}{{\text{csch}\,y\coth y}} } = {\frac{1}{{x\sqrt {1 + {x^2}} }}\;\;}\kern-0.3pt{\left( {x \lt 0} \right).}$

By combining the two branches of the solutions, we obtain the final expression for the derivative of the inverse hyperbolic cosecant in the form

${\left( {\text{arccsch}\,x} \right)^\prime } = { – \frac{1}{{\left| x \right|\sqrt {1 + {x^2}} }}\;\;}\kern-0.3pt{\left( {x \ne 0} \right).}$

### Table of Derivatives of Hyperbolic Functions

For convenience, we collect the differentiation formulas for all hyperbolic functions in one table:

In the examples below, find the derivative of the given function.

## Solved Problems

Click on problem description to see solution.

### ✓Example 1

$y = \coth \frac{1}{x}$

### ✓Example 2

${y = \ln \left( {\sinh x} \right),\;\;}\kern-0.3pt{x \gt 0.}$

### ✓Example 3

$y = \sinh \left( {\tan x} \right)$

### ✓Example 4

$y = \tanh \left( {{x^2}} \right)$

### ✓Example 5

$y = x\sinh x – \cosh x$

### ✓Example 6

$y = {\sinh ^2}x$

### ✓Example 7

$y = \sinh x\tanh x$

### ✓Example 8

$y = \text{arctanh}\frac{1}{x}$

### ✓Example 9

$y = \text{arctanh}\left( {\cos x} \right)$

### ✓Example 10

$y = \text{arccosh}\frac{x}{a}$

### ✓Example 11

$y = {\text{csch}^2}\left( {3x} \right)$

### ✓Example 12

$y = \text{arcsinh}\left( {\tan x} \right)$

### ✓Example 13

$y = {\text{sech}^2}\ln x$

### ✓Example 14

Prove the identity $${\text{arcsinh}\,x }={ \ln \left( {x + \sqrt {1 + {x^2}} } \right).}$$

### ✓Example 15

$y = \text{arccosh}\frac{1}{{{x^2}}}$

### ✓Example 16

$y = \text{arctanh}\left( {2\sqrt x } \right)$

### ✓Example 17

$y = \arctan \left( {\tanh x} \right)$

### ✓Example 18

$y = \ln \left( {\cosh x} \right) + \frac{1}{{2{{\cosh }^2}x}}$

### ✓Example 19

$y = \arccos \left( {\frac{1}{{\cosh x}}} \right)$

### ✓Example 20

$y = \text{arcsech}\sqrt {1 – {x^2}}$

### Example 1.

$y = \coth \frac{1}{x}$

#### Solution.

Differentiating as a composite function, we find:

${y’\left( x \right) }={ {\left( {\coth \frac{1}{x}} \right)^\prime } } = { – {\text{csch}^2}\left( {\frac{1}{x}} \right) \cdot {\left( {\frac{1}{x}} \right)^\prime } } = { – {\text{csch}^2}\left( {\frac{1}{x}} \right) \cdot \left( { – \frac{1}{{{x^2}}}} \right) } = {\frac{{{{\text{csch}}^2}\left( {\frac{1}{x}} \right)}}{{{x^2}}}.}$

### Example 2.

${y = \ln \left( {\sinh x} \right),\;\;}\kern-0.3pt{x \gt 0.}$

#### Solution.

${y’\left( x \right) }={ {\left[ {\ln \left( {\sinh x} \right)} \right]^\prime } } = {\frac{1}{{\sinh x}} \cdot {\left( {\sinh x} \right)^\prime } } = {\frac{{\cosh x}}{{\sinh x}} = \tanh x.}$

### Example 3.

$y = \sinh \left( {\tan x} \right)$

#### Solution.

Using the chain rule, we obtain:

${y’\left( x \right) }={ {\left[ {\sinh \left( {\tan x} \right)} \right]^\prime } } = {\cosh \left( {\tan x} \right) \cdot {\left( {\tan x} \right)^\prime } } = {\cosh \left( {\tan x} \right) \cdot \frac{1}{{{{\cos }^2}x}} } = {\frac{{\cosh \left( {\tan x} \right)}}{{{{\cos }^2}x}},}$

where $$x \ne {\large\frac{\pi }{2}\normalsize} + \pi n,$$ $$n \in \mathbb{Z}.$$

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Problems 1-3
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Problems 4-20