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