Derivatives of Hyperbolic Functions
Solved Problems
Example 15.
\[y = {\text{csch}^2}\left( {3x} \right)\]
Solution.
Using the chain rule twice, we obtain:
\[y'\left( x \right) = \left[ {{{\text{csch}}^2}\left( {3x} \right)} \right]^\prime = 2\,\text{csch}\left( {3x} \right) \cdot {\left[ {\text{csch}\left( {3x} \right)} \right]^\prime } = 2\,\text{csch}\left( {3x} \right) \cdot \left[ { - \text{csch}\left( {3x} \right)\coth \left( {3x} \right)} \right] \cdot {\left( {3x} \right)^\prime } = - 2\,\text{csch}\left( {3x} \right) \cdot \text{csch}\left( {3x} \right) \cdot \coth \left( {3x} \right) \cdot 3 = - 6\,{\text{csch}^2}\left( {3x} \right) \coth \left( {3x} \right)\;\;\left( {x \ne 0} \right).\]
Example 16.
\[y = \text{arcsinh}\left( {\tan x} \right)\]
Solution.
Applying the chain rule, we get:
\[y'\left( x \right) = \left[ {\text{arcsinh}\left( {\tan x} \right)} \right]^\prime = \frac{1}{{\sqrt {{{\tan }^2}x + 1} }} \cdot {\left( {\tan x} \right)^\prime } = \frac{1}{{\sqrt {{{\tan }^2}x + 1} }} \cdot \frac{1}{{{{\cos }^2}x}}.\]
This expression can be simplified using the trigonometric identity
\[{\tan ^2}x + 1 = \frac{1}{{{{\cos }^2}x}}.\]
Hence,
\[y'\left( x \right) = \frac{1}{{\sqrt {{{\tan }^2}x + 1} }} \cdot \frac{1}{{{{\cos }^2}x}} = \frac{1}{{\sqrt {\frac{1}{{{{\cos }^2}x}}} }} \cdot \frac{1}{{{{\cos }^2}x}} = \frac{{\cancel{\cos x}}}{{{{\cos }^{\cancel{2}}}x}} = \frac{1}{{\cos x}} = \sec x,\]
where \(x \ne {\frac{\pi }{2}} + \pi n,\) \(n \in \mathbb{Z}.\)
Example 17.
\[y = {\text{sech}^2}\ln x\]
Solution.
Similarly, by the chain rule, we can write:
\[y'\left( x \right) = {\left( {{{\text{sech}}^2}\ln x} \right)^\prime } = 2\,\text{sech}\ln x \cdot {\left( {\text{sech}\ln x} \right)^\prime } = 2\,\text{sech}\ln x \cdot \left( { - \text{sech}\ln x \cdot \tanh \ln x} \right) \cdot {\left( {\ln x} \right)^\prime } = - \frac{2}{x}{\text{sech}^2}\ln x \cdot \tanh \ln x.\]
Example 18.
Prove the identity \(\text{arcsinh}\,x =\) \(\ln \left( {x + \sqrt {1 + {x^2}} } \right).\)
Solution.
We differentiate both sides of this expression and then simplify:
\[\left( {\text{arcsinh}\,x} \right)^\prime = \left[ {\ln \left( {x + \sqrt {1 + {x^2}} } \right)} \right]^\prime,\;\; \Rightarrow \frac{1}{{\sqrt {{x^2} + 1} }} = \frac{1}{{x + \sqrt {1 + {x^2}} }} \cdot {\left( {x + \sqrt {1 + {x^2}} } \right)^\prime },\;\; \Rightarrow \frac{1}{{\sqrt {{x^2} + 1} }} = \frac{{1 + \frac{1}{{2\sqrt {1 + {x^2}} }} \cdot 2x}}{{x + \sqrt {1 + {x^2}} }},\;\; \Rightarrow \frac{1}{{\sqrt {{x^2} + 1} }} = \frac{{\frac{\cancel{x + \sqrt {1 + {x^2}} }}{{\sqrt {1 + {x^2}} }}}}{\cancel{x + \sqrt {1 + {x^2}} }},\;\; \Rightarrow \frac{1}{{\sqrt {{x^2} + 1} }} \equiv \frac{1}{{\sqrt {{x^2} + 1} }}.\]
Hence, the initial formula is proved (at least with the accuracy of a constant).
Example 19.
\[y = \text{arccosh}\frac{1}{{{x^2}}}\]
Solution.
Using the chain rule, we can write:
\[y'\left( x \right) = \left( {\text{arccosh}\frac{1}{{{x^2}}}} \right)^\prime = \frac{1}{{\sqrt {{{\left( {\frac{1}{{{x^2}}}} \right)}^2} - 1} }} \cdot {\left( {\frac{1}{{{x^2}}}} \right)^\prime } = \frac{1}{{\sqrt {\frac{{1 - {x^4}}}{{{x^4}}}} }} \cdot \left( { - \frac{2}{{{x^3}}}} \right) = - \frac{{2x}}{{{x^3}\sqrt {1 - {x^4}} }} = - \frac{2}{{x\sqrt {1 - {x^4}} }}.\]
The given function and its derivative exist for all \(x\) satisfying the inequality \({x^2} \lt 1.\)
Example 20.
\[y = \text{arctanh}\left( {2\sqrt x } \right)\]
Solution.
By the chain rule,
\[y'\left( x \right) = \left[ {\text{arctanh}\left( {2\sqrt x } \right)} \right]^\prime = \frac{1}{{1 - {{\left( {2\sqrt x } \right)}^2}}} \cdot {\left( {2\sqrt x } \right)^\prime } = \frac{1}{{1 - 4x}} \cdot \frac{\cancel{2}}{{\cancel{2}\sqrt x }} = \frac{1}{{\sqrt x \left( {1 - 4x} \right)}}.\]
This answer is correct if
\[\left\{ \begin{array}{l} \left| {2\sqrt x } \right| \lt 1\\ x \gt 0 \end{array} \right.\;\;\text{or}\;\;0 \lt x \lt \frac{1}{4}. \]
Example 21.
\[y = \arctan \left( {\tanh x} \right)\]
Solution.
\[y'\left( x \right) = {\left[ {\arctan \left( {\tanh x} \right)} \right]^\prime } = \frac{1}{{1 + {{\tanh }^2}x}} \cdot {\left( {\tanh x} \right)^\prime } = \frac{1}{{1 + {{\tanh }^2}x}} \cdot \frac{1}{{{{\cosh }^2}x}} = \frac{1}{{\left( {1 + \frac{{{{\sinh }^2}x}}{{{{\cosh }^2}x}}} \right){{\cosh }^2}x}} = \frac{1}{{{{\cosh }^2}x + {{\sinh }^2}x}}.\]
The denominator can be simplified by the double angle formula
\[\cosh 2x = {\cosh ^2}x + {\sinh ^2}x.\]
As a result, we obtain the answer in the form
\[y'\left( x \right) = \frac{1}{{{{\cosh }^2}x + {{\sinh }^2}x}} = \frac{1}{{\cosh 2x}}.\]
Example 22.
\[y = \ln \left( {\coth {\frac{x}{2}}} \right)\]
Solution.
Using the chain rule, we can write:
\[y^\prime = \left[ {\ln \left( {\coth \frac{x}{2}} \right)} \right]^\prime = \frac{1}{{\coth \frac{x}{2}}} \cdot \left( {\coth \frac{x}{2}} \right)^\prime = \tanh \frac{x}{2} \cdot \left( { - \frac{1}{{{{\sinh }^2}\frac{x}{2}}}} \right) \cdot \left( {\frac{x}{2}} \right)^\prime = - \frac{{\sinh \frac{x}{2}}}{{\cosh \frac{x}{2}}} \cdot \frac{1}{{{{\sinh }^2}\frac{x}{2}}} \cdot \frac{1}{2} = - \frac{{\cancel{\sinh \frac{x}{2}}}}{{2\cosh \frac{x}{2}{{\sinh }^\cancel{2}}\frac{x}{2}}} = - \frac{1}{{2\cosh \frac{x}{2}\sinh \frac{x}{2}}}.\]
Given the hyperbolic identity
\[\sinh 2x = 2\sinh x\cosh x,\]
we get the derivative in the form
\[y^\prime = - \frac{1}{{2\cosh \frac{x}{2}\sinh \frac{x}{2}}} = - \frac{1}{{\sinh x}} = - \text{csch}\,x.\]
Example 23.
\[y = \ln \left( {\cosh x} \right) + \frac{1}{{2{{\cosh }^2}x}}\]
Solution.
Using the linear properties of the derivative and the chain rule, we have
\[y'\left( x \right) = \left[ {\ln \left( {\cosh x} \right) + \frac{1}{{2{{\cosh }^2}x}}} \right]^\prime = \left( {\ln \left( {\cosh x} \right)} \right)^\prime + \left( {\frac{1}{{2{{\cosh }^2}x}}} \right)^\prime = \frac{{\sinh x}}{{\cosh x}} - \frac{{\cancel{2}\sinh x}}{{\cancel{2}{{\cosh }^3}x}} = \tanh x - \tanh x \cdot \frac{1}{{{{\cosh }^2}x}} = \tanh x\left( {1 - \frac{1}{{{{\cosh }^2}x}}} \right).\]
Using the relationship
\[1 - \frac{1}{{{{\cosh }^2}x}} = {\tanh ^2}x,\]
which follows from the identity
\[{\cosh ^2}x - {\sinh ^2}x = 1,\]
we get the following answer:
\[y'\left( x \right) = \tanh x\left( {1 - \frac{1}{{{{\cosh }^2}x}}} \right) = \tanh x \cdot {\tanh ^2}x = {\tanh ^3}x.\]
Example 24.
\[y = \arccos \left( {\frac{1}{{\cosh x}}} \right)\]
Solution.
Applying the chain rule twice, we obtain:
\[y'\left( x \right) = \left[ {\arccos \left( {\frac{1}{{\cosh x}}} \right)} \right]^\prime = - \frac{1}{{\sqrt {1 - {{\left( {\frac{1}{{\cosh x}}} \right)}^2}} }} \cdot {\left( {\frac{1}{{\cosh x}}} \right)^\prime } = - \frac{1}{{\sqrt {1 - \frac{1}{{{{\cosh }^2}x}}} }} \cdot \left( { - \frac{1}{{{{\cosh }^2}x}}} \right) \cdot {\left( {\cosh x} \right)^\prime } = \frac{1}{{\sqrt {\frac{{{{\cosh }^2}x - 1}}{{{{\cosh }^2}x}}} }} \cdot \frac{{\sinh x}}{{{{\cosh }^2}x}} = \frac{{\cosh x}}{{\sqrt {{{\cosh }^2}x - 1} }} \cdot \frac{{\sinh x}}{{{{\cosh }^2}x}}.\]
Take into account that \({\cosh ^2}x - 1 = {\sinh ^2}x.\) Hence,
\[y'\left( x \right) = \frac{{\cancel{\cosh x} \cdot \sinh x}}{{{{\cosh }^{\cancel{2}}}x\sqrt {{{\cosh }^2}x - 1} }} = \frac{{\sinh x}}{{\cosh x\sqrt {{\sinh^2}x} }} = \frac{{\sinh x}}{{\cosh x\left| {\sinh x} \right|}}.\]
The ratio \(\frac{{\sinh x}}{{\left| {\sinh x} \right|}}\) is equal to \(\pm 1\) depending on the sign of the argument \(x.\) Therefore, the final answer can be written as
\[y'\left( x \right) = \frac{{\sinh x}}{{\cosh x\left| {\sinh x} \right|}} = \frac{{\text{sign}\,x}}{{\cosh x}}\;\;\left( {x \ne 0} \right).\]
Example 25.
\[y = \text{arcsech}{\frac{1}{x}}\]
Solution.
Using the chain rule, we obtain:
\[y^\prime = \left( {\text{arcsech} \frac{1}{x}} \right)^\prime = - \frac{1}{{\frac{1}{x}\sqrt {1 - {{\left( {\frac{1}{x}} \right)}^2}} }} \cdot \left( {\frac{1}{x}} \right)^\prime = - \frac{x}{{\sqrt {1 - \frac{1}{{{x^2}}}} }} \cdot \left( { - \frac{1}{{{x^2}}}} \right) = \frac{1}{{x\sqrt {\frac{{{x^2} - 1}}{{{x^2}}}} }} = \frac{{\sqrt {{x^2}} }}{{x\sqrt {{x^2} - 1} }} = \frac{{\left| x \right|}}{{x\sqrt {{x^2} - 1} }}.\]
The function is defined for
\[0 \lt \frac{1}{x} \lt 1,\;\; \Rightarrow x \gt 1.\]
Then
\[\frac{{\left| x \right|}}{x} = \frac{\cancel{x}}{\cancel{x}} = 1,\]
so the derivative is given by
\[y^\prime = \frac{1}{{\sqrt {{x^2} - 1} }}.\]
Example 26.
\[y = \text{arcsech}\sqrt {1 - {x^2}} \]
Solution.
First we investigate the domain of this function:
\[ 0 \lt \sqrt {1 - {x^2}} \lt 1,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{c}} {\sqrt {1 - {x^2}} \gt 0}\\ {\sqrt {1 - {x^2}} \lt 1} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{c}} {1 - {x^2} \gt 0}\\ {1 - {x^2} \lt 1} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{c}} {{x^2} \lt 1}\\ {{x^2} \gt 0} \end{array}} \right.,\;\; \Rightarrow \left| x \right| \lt 1,\;\;x \ne 0.\]
Then we calculate the derivative using the chain rule:
\[y'\left( x \right) = \left( {\text{arcsech}\sqrt {1 - {x^2}} } \right)^\prime = - \frac{1}{{\sqrt {1 - {x^2}} \sqrt {1 - {{\left( {\sqrt {1 - {x^2}} } \right)}^2}} }} \cdot {\left( {\sqrt {1 - {x^2}} } \right)^\prime } = - \frac{1}{{\sqrt {1 - {x^2}} \sqrt {1 - \left( {1 - {x^2}} \right)} }} \cdot \frac{1}{{2\sqrt {1 - {x^2}} }} \cdot \left( { - 2x} \right) = \frac{x}{{{{\left( {\sqrt {1 - {x^2}} } \right)}^2}\sqrt {{x^2}} }} = \frac{x}{{\left| x \right|\left( {1 - {x^2}} \right)}} = \frac{{\text{sign}\,x}}{{1 - {x^2}}}\;\;\left( {\left| x \right| \lt 1,\;x \ne 0} \right).\]
