# Derivatives of Hyperbolic Functions

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

## Solved Problems

### 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.

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

Solution.

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

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

### Example 4.

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

Solution.

By the chain rule,

$y^\prime = \left[ {\sinh \left( {\ln x} \right)} \right]^\prime = \cosh \left( {\ln x} \right) \cdot \left( {\ln x} \right)^\prime = \cosh \left( {\ln x} \right) \cdot \frac{1}{x} = \frac{{{e^{\ln x}} + {e^{ - \ln x}}}}{{2x}}.$

We simplify the numerator using the logarithmic identity:

$e^{\ln x} = x;$
${e^{ - \ln x}} = e^{\ln {x^{ - 1}}} = {e^{\ln \frac{1}{x}}} = \frac{1}{x}.$

Hence,

$y^\prime = \frac{{{e^{\ln x}} + {e^{ - \ln x}}}}{{2x}} = \frac{{x + \frac{1}{x}}}{{2x}} = \frac{{{x^2} + 1}}{{2{x^2}}}.$

### Example 5.

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

Solution.

By the chain rule, we have:

$y'\left( x \right) = \left[ {\tanh \left( {{x^2}} \right)} \right]^\prime = \frac{1}{{{{\cosh }^2}\left( {{x^2}} \right)}} \cdot {\left( {{x^2}} \right)^\prime } = \frac{{2x}}{{{{\cosh }^2}\left( {{x^2}} \right)}}.$

### Example 6.

$y = x\sinh x - \cosh x$

Solution.

Using the difference and product rule, we have:

$\require{cancel} y'\left( x \right) = {\left( {x\sinh x - \cosh x} \right)^\prime } = {\left( {x\sinh x} \right)^\prime } - {\left( {\cosh x} \right)^\prime } = x'\sinh x + x{\left( {\sinh x} \right)^\prime } - {\left( {\cosh x} \right)^\prime } = 1 \cdot \sinh x + x \cdot \cosh x - \sinh x = \cancel{\sinh x} + x\cosh x - \cancel{\sinh x} = x\cosh x.$

### Example 7.

$y = \sinh x\cosh x - x$

Solution.

Using the hyperbolic identity

$\sinh 2x = 2\sinh x\cosh x,$

we can write the equation in the form

$y = \sinh x\cosh x - x = \frac{1}{2}\sinh 2x - x.$

Applying the chain rule, we have

$y^\prime = \left( {\frac{1}{2}\sinh 2x - x} \right)^\prime = \frac{1}{2}\cosh 2x \cdot \left( {2x} \right)^\prime - 1 = \frac{1}{2}\cosh 2x \cdot 2 - 1 = \cosh 2x - 1.$

Now we use another hyperbolic identity

$\cosh 2x - 1 = 2\,{\sinh ^2}x.$

So

$y^\prime = 2\,{\sinh ^2}x.$

### Example 8.

$y = {\sinh ^2}x$

Solution.

$y'\left( x \right) = \left( {{{\sinh }^2}x} \right)^\prime = 2\sinh x \cdot {\left( {\sinh x} \right)^\prime } = 2\sinh x\cosh x.$

We can simplify the answer using the double angle identity $$\sinh 2x$$ $$= 2\sinh x\cosh x$$. Hence,

$y'\left( x \right) = 2\sinh x\cosh x = \sinh 2x.$

### Example 9.

$y = \sinh x\tanh x$

Solution.

Using the product rule for differentiation, we obtain:

$y'\left( x \right) = \left( {\sinh x\tanh x} \right)^\prime = {\left( {\sinh x} \right)^\prime }\tanh x + \sinh x{\left( {\tanh x} \right)^\prime } = \cosh x \cdot \tanh x + \sinh x \cdot \frac{1}{{{{\cosh }^2}x}} = \frac{{\cancel{\cosh x}\sinh x}}{{\cancel{\cosh x}}} + \sinh x\,{\text{sech}^2}x = \sinh x\left( {1 + {{\text{sech}}^2}x} \right).$

### Example 10.

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

Solution.

$y'\left( x \right) = \left( {\text{arctanh}\frac{1}{x}} \right)^\prime = \frac{1}{{1 - {{\left( {\frac{1}{x}} \right)}^2}}} \cdot {\left( {\frac{1}{x}} \right)^\prime } = \frac{1}{{1 - \frac{1}{{{x^2}}}}} \cdot \left( { - \frac{1}{{{x^2}}}} \right) = - \frac{{{x^2}}}{{{x^2} - 1}} \cdot \frac{1}{{{x^2}}} = \frac{1}{{1 - {x^2}}}.$

Interestingly, that the derivatives of the functions $$y = \text{arctanh}{\frac{1}{x}}$$ and $$y = \text{arctanh}\,x$$ are the same.

### Example 11.

$y = \text{arctanh} \frac{1}{{{x^2}}}, \;\left| x \right| \lt 1.$

Solution.

We use the derivative of the inverse tangent hyperbolic function along with the chain rule:

$y^\prime = \left( {\text{arctanh}\frac{1}{{{x^2}}}} \right)^\prime = \frac{1}{{1 - {{\left( {\frac{1}{{{x^2}}}} \right)}^2}}} \cdot \left( {\frac{1}{{{x^2}}}} \right)^\prime = \frac{1}{{1 - \frac{1}{{{x^4}}}}} \cdot \left( { - 2{x^{ - 3}}} \right) = - \frac{{2{x^4}}}{{\left( {{x^4} - 1} \right){x^3}}} = - \frac{{2x}}{{{x^4} - 1}} = \frac{{2x}}{{1 - {x^4}}}.$

### Example 12.

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

Solution.

$y'\left( x \right) = \left[ {\text{arctanh}\left( {\cos x} \right)} \right]^\prime = \frac{1}{{1 - {{\cos }^2}x}} \cdot {\left( {\cos x} \right)^\prime } = \frac{1}{{{{\sin }^2}x}} \cdot \left( { -\sin x} \right) = - \frac{{\cancel{\sin x}}}{{{{\sin }^{\cancel{2}}}x}} = - \frac{1}{{\sin x}} = - \csc x.$

The domain of $$x$$ is determined by the inequality $$x \ne {\frac{\pi }{2}} + \pi n,$$ $$n \in \mathbb{Z}.$$

### Example 13.

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

Solution.

$y'\left( x \right) = \left( {\text{arccosh}\frac{x}{a}} \right)^\prime = \frac{1}{{\sqrt {{{\left( {\frac{x}{a}} \right)}^2} - 1} }} \cdot {\left( {\frac{x}{a}} \right)^\prime } = \frac{1}{{\sqrt {\frac{{{x^2} - {a^2}}}{{{a^2}}}} }} \cdot \frac{1}{a} = \frac{\cancel{a}}{{\cancel{a}\sqrt {{x^2} - {a^2}} }} = \frac{1}{{\sqrt {{x^2} - {a^2}} }}.$

The solution exists for $${\frac{x}{a}} \gt 1.$$

### Example 14.

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

Solution.

Using the chain rule twice, we obtain:

$y^\prime = \left( {\text{arccosh} \frac{a}{x}} \right)^\prime = \frac{1}{{\sqrt {{{\left( {\frac{a}{x}} \right)}^2} - 1} }} \cdot \left( {\frac{a}{x}} \right)^\prime = \frac{1}{{\sqrt {\frac{{{a^2}}}{{{x^2}}} - 1} }} \cdot \left( { - \frac{a}{{{x^2}}}} \right) = - \frac{{a\sqrt {{x^2}} }}{{{x^2}\sqrt {{a^2} - {x^2}} }} = - \frac{{a\left| x \right|}}{{{x^2}\sqrt {{a^2} - {x^2}} }}.$

The domain of the function is determined by the condition

$\frac{a}{x} \gt 1,\;\; \Rightarrow 0 \lt x \lt a.$

Then $$\left| x \right| = x$$ and the derivative is given by

$y^\prime = - \frac{a}{{x\sqrt {{a^2} - {x^2}} }}.$