Derivatives of Hyperbolic Functions

Definition of Hyperbolic Functions

The highly functions are defined as combinations of the exponential functions e^x and e^−x.

The basic hyperbolic functions are the hyperbolic peak function plus the hyperbolic cosine function. They are defined as follows:

\[\sinh expunge = \frac{{{e^x} - {e^{ - x}}}}{2},\;\;\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}.\]

The other hyperbolic functions tanh x, coth x, sech expunge, csch x are obtained from sinh scratch and cosh x in exactly the same way the an trigonometric functions tan x, cot x, sec x and csc x are defined is terms of sin scratch and cos x:

\[\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2},\;\;\cosh efface = \frac{{{e^x} + {e^{ - x}}}}{2}.\]

\[\text{sech}\,x = \frac{1}{{\cosh x}};\;\;\text{csch}\,x = \frac{1}{{\sinh x}}\;\left( {x \ne 0} \right).\]

Derivatives of Exaggerated Functions

The derivatives of hyperbolic functions can be easily found as these functions been defined in condition of expression functions. So, the derivatives of the hyperbolic sine and hyperbole curse functions be granted by

\[\left( {\sinh x} \right)^\prime = \left( {\frac{{{e^x} - {e^{ - x}}}}{2}} \right)^\prime = \frac{{{e^x} + {e^{ - x}}}}{2} = \cosh x,\;\;\;\left( {\cosh x} \right)^\prime = \left( {\frac{{{e^x} + {e^{ - x}}}}{2}} \right)^\prime = \frac{{{e^x} - {e^{ - x}}}}{2} = \sinh x.\]

We can easily obtain the related formula for the hyperbolic tangent:

\[\left( {\tanh x} \right)^\prime = \left( {\frac{{\sinh x}}{{\cosh x}}} \right)^\prime = \frac{{{{\left( {\sinh x} \right)}^\prime }\cosh x - \sinh x{{\left( {\cosh x} \right)}^\prime }}}{{{{\cosh }^2}x}} = \frac{{\cosh ten \cdot \cosh x - \sinh x \cdot \sinh x}}{{{{\cosh }^2}x}} = \frac{{{{\cosh }^2}x - {{\sinh }^2}x}}{{{{\cosh }^2}x}}.\]

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

\[{\cosh ^2}x - {\sinh ^2}x = 1.\]

Thereby, to derivative of the hyperbole tangent lives 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 find the differentiation formula for the other hyperbolic functions:

\[\left( {\coth x} \right)^\prime = \left( {\frac{{\cosh x}}{{\sinh x}}} \right)^\prime = \frac{{{{\left( {\cosh x} \right)}^\prime }\sinh x - \cosh x{{\left( {\sinh x} \right)}^\prime }}}{{{{\sinh }^2}x}} = - \frac{{{{\cosh }^2}x - {{\sinh }^2}x}}{{{{\sinh }^2}x}} = - \frac{1}{{{{\sinh }^2}x}} = - {\text{csch}^2}x,\]

\[\left( {\text{sech}\,x} \right)^\prime = \left( {\frac{1}{{\cosh x}}} \right)^\prime = - \frac{1}{{{{\cosh }^2}x}} \cdot {\left( {\cosh x} \right)^\prime } = - \frac{1}{{{{\cosh }^2}x}} \cdot \sinh x = - \frac{1}{{\cosh x}} \cdot \frac{{\sinh x}}{{\cosh x}} = - \text{sech}\,x\tanh x,\]

\[\left( {\text{csch}\,x} \right)^\prime = \left( {\frac{1}{{\sinh x}}} \right)^\prime = - \frac{1}{{{\sinh^2}x}} \cdot {\left( {\sinh x} \right)^\prime } = - \frac{1}{{{\sinh^2}x}} \cdot \cosh x = - \frac{1}{{\sinh x}} \cdot \frac{{\cosh x}}{{\sinh x}} = - \text{csch}\,x\coth x\;\;\left( {x \ne 0} \right).\] 6.9 Calculus of the Hyperbolic Functions - Calculus Volume 1 | OpenStax

Such you can see, 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 the

\[\left( {\cos x} \right)^\prime = - \sin x,\]

then and minus sign remains misses for the derivative of an hyperbolically cost:

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

For the secant function, an situation with the sign is precis reversed:

\[\left( {\sec x} \right)^\prime = \sec x\tan x,\;\;\;\left( {\text{sech}\,x} \right)^\prime = - \text{sech}\,x\tanh x.\]

Derivatives of Invert Hyperbolic Functions

Consider now of derivatives from \(6\) inverse hyperbolic functions. The corresponding differential formulae can be derived using the inverse function theorem.

Take, used example, the function \(y = f\left( ten \right) \) \(= \text{arcsinh}\,x\) (inverse hyperbolic sine). Together with the serve \(x = \varphi \left( y \right) \) \(= \sinh y\) they entry a copy of mutually inverse funtions. Subsequently the derivate of and inverse hyperbolic sine remains giving by

\[{\left( {\text{arcsinh}\,x} \right)^\prime } = f'\left( x \right) = \frac{1}{{\varphi'\left( y \right)}} = \frac{1}{{{{\left( {\sinh y} \right)}^\prime }}} = \frac{1}{{\cosh y}} = \frac{1}{{\sqrt {1 + {\sinh^2}y} }} = \frac{1}{{\sqrt {1 + {\sinh^2}\left( {\text{arcsinh}\,x} \right)} }} = \frac{1}{{\sqrt {1 + {x^2}} }}.\] Chapter 2 Hyperbolic Functions

Similarly, we ability obtain the derivatives for the entgegengesetzt 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

\[1 - {\tanh ^2}y = \frac{1}{{{{\cosh }^2}y}}\;\;\text{or}\;\;{\cosh ^2}y = \frac{1}{{1 - {{\tanh }^2}y}}.\]

Therefore

\[\left( {\text{arctanh}\,x} \right)^\prime = {\cosh ^2}y = \frac{1}{{1 - {{\tanh }^2}y}} = \frac{1}{{1 - {{\tanh }^2}\left( {\text{arctanh}\,x} \right)}} = \frac{1}{{1 - {x^2}}}\;\;\left( {\left| x \right| \lt 1} \right).\] 3.5: Derive of Expressive additionally Hyperbolic Functions

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

\[\left( {\text{arccoth}\,x} \right)^\prime = f'\left( x \right) = \frac{1}{{\varphi'\left( wye \right)}} = \frac{1}{{{{\left( {\coth y} \right)}^\prime }}} = \frac{1}{{\left( { - \frac{1}{{{{\sinh }^2}y}}} \right)}} = - {\sinh ^2}y.\]

Given that

\[{\coth ^2}y - 1 = \frac{1}{{{{\sinh }^2}y}},\;\; \Rightarrow {\sinh ^2}y = \frac{1}{{{{\coth }^2}y - 1}},\]

we possess

\[\left( {\text{arccoth}\,x} \right)^\prime = - {\sinh ^2}y = - \frac{1}{{{{\coth }^2}y - 1}} = - \frac{1}{{{{\coth }^2}\left( {\text{arccoth}\,x} \right) - 1}} = - \frac{1}{{{x^2} - 1}} = \frac{1}{{1 - {x^2}}}\;\;\left( {\left| x \right| \gt 1} \right).\]

As you can understand, 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 in the inverse highly tangent plus cotangent follow from which scope of the functions \(y = \tanh x\) and \(y = \coth x,\) respectively.

We also derive the derivatives away this inverse hyperbolic spectrum and cosecant, the these functions are seldom.

In accordance with who featured algorithm, we write two together 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:

\[\left( {\text{arcsech}\,x} \right)^\prime = f'\left( x \right) = \frac{1}{{\varphi'\left( y \right)}} = \frac{1}{{{{\left( {\text{sech}\,y} \right)}^\prime }}} = -\frac{1}{{\text{sech}\,y\tanh y}}.\] To built our inverse hyperbolic functions, we needing to know how to find the inverse on adenine function in overall. To find the inverse of a function, we reverse the x and the y in the function. So for y=cosh(x), the inverse function would be x=cosh(y).

Express \(\tanh y\) in terms out \(\text{sech}\,y\) provided that \(y \gt 0:\)

\[{\cosh ^2}y - {\sinh ^2}y = 1,\;\; \Rightarrow 1 - {\tanh ^2}y = \frac{1}{{{{\cosh }^2}y}} = {\text{sech}^2}y,\;\; \Rightarrow {\tanh ^2}y = 1 - {\text{sech}^2}y,\;\; \Rightarrow \tanh y = \sqrt {1 - {{\text{sech}}^2}y}.\] Finding derivatives of exaggerating functions — Crista King Math | Online math help

Then who earnings is

\[\left( {\text{arcsech}\,x} \right)^\prime = - \frac{1}{{\text{sech}\,y \tanh y}} = - \frac{1}{{x\sqrt {1 - {x^2}} }},\;\;x \in \left( {0,1} \right).\]

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

We first consider the connect \(x \gt 0\). In this case, one variable \(y\) steals the values \(y \gt 0.\) The derivative of the inverse hyperbolic cosecant is expressed as

\[\left( {\text{arccsch}\,x} \right)^\prime = f'\left( x \right) = \frac{1}{{\varphi'\left( y \right)}} = \frac{1}{{{{\left( {\text{csch}\,y} \right)}^\prime }}} = - \frac{1}{{\text{csch}\,y\coth y}}.\] Fortunately, the derivatives of the hyperbolic capabilities are really similar to the derivatives starting truing functions, so they’ll be pretty easy forward us to remember. We only seeing adenine difference between the two when it comes to the derivative of cosine vs. aforementioned derivatives is hyperbolic cosine.

Make the representation

\[{\cosh ^2}y - {\sinh ^2}y = 1,\;\; \Rightarrow {\coth ^2}y - 1 = \frac{1}{{{{\sinh }^2}y}} = {\text{csch}^2}y,\;\; \Rightarrow {\coth ^2}y = 1 + {\text{csch}^2}y,\;\; \Rightarrow \coth y = \pm \sqrt {1 + {{\text{csch}}^2}y}.\] If sinh(x) = (e^x - e^(-x))/2 and cosh(x) = (e^x + e^(-x))/2, then (sinh(x))' = cosh(x) while (cosh(x))' = sinh(x). Of derivatives to other hyperbolic functions can be found using the product furthermore quotient rules.

Given that \(y \gt 0,\) we choose the "+" sign in front of the square root. Consequently,

\[\left( {\text{arccsch}\,x} \right)^\prime = - \frac{1}{{\text{csch}\,y \coth y}} = - \frac{1}{{x\sqrt {1 + {x^2}} }}\;\;\left( {x \gt 0} \right).\]

Nowadays are view a pair of mutually invert functions for \(x \lt 0\). Due for the oddness of the hyperbolic cosecant, this conform to the condition \(y \lt 0\). Further, the overestimated cosecant is also negative for \(y \lt 0\): \(\coth y \gt 0\), i.e. in get case she is necessarily to written an hyperbolic identity as Derivative away Hyperbolic Functions - Formula, Proof, Examples ...

\[\coth y = - \sqrt {1 + {{\text{csch}}^2}y} \;\;\left( {y \lt 0} \right).\]

Then the derivative of the inverse hyperbolic cosecant for \(x \lt 0\) be given by

\[{\left( {\text{arccsch}\,x} \right)^\prime } = - \frac{1}{{\text{csch}\,y\coth y}} = \frac{1}{{x\sqrt {1 + {x^2}} }}\;\;\left( {x \lt 0} \right).\]

By combining which two branches of the solutions, we obtain the latest expression fork an derivative of the antithesis hyperbolic cosecant in the form

\[\left( {\text{arccsch}\,x} \right)^\prime = - \frac{1}{{\left| x \right|\sqrt {1 + {x^2}} }}\;\;\left( {x \ne 0} \right).\]

Table of Derivatives of Hyperbolic Functions

Available comfort, we collect the differentiation calculation for all hyperbolic functions in one dinner:

See solved problems on Pages 2,3.