In mathematics, when $x$ represents a variable, the inverse hyperbolic cotangent function is written as $\coth^{-1}{x}$ or $\operatorname{arccoth}{x}$. The differentiation or the derivative of inverse hyperbolic cot function with respect to $x$ is expressed in two different following forms.
$(1).\,\,\,$ $\dfrac{d}{dx}{\, \Big(\coth^{-1}{(x)}\Big)}$
$(2).\,\,\,$ $\dfrac{d}{dx}{\, \Big(\operatorname{arccoth}{(x)}\Big)}$
The derivative formula for the inverse hyperbolic cotangent function can be derived in mathematics from the first principle of differentiation. So, let us learn how to derive the differentiation formula for the inverse hyperbolic cot function in differential calculus.
According to the fundamental definition of the derivative, the derivative of the inverse hyperbolic co-tangent function can be proved in limit form.
$\dfrac{d}{dx}{\, (\coth^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to \, 0}{\normalsize \dfrac{\coth^{-1}{(x+\Delta x)}-\coth^{-1}{x}}{\Delta x}}$
Let the differential element $\Delta x$ is represented by $h$ simply, then the whole expression can be written in $h$ instead of $\Delta x$ for our convenience.
$\implies$ $\dfrac{d}{dx}{\, (\coth^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\coth^{-1}{(x+h)}-\coth^{-1}{x}}{h}}$
Now, the differentiation rule for the $\operatorname{arccoth}{(x)}$ function with respect to $x$ is derived in differential calculus from the first principle of differentiation.
The direct substitution method is a fundamental approach of evaluating the limit of any mathematical expression in calculus. Hence, use it for deriving the differentiation of inverse hyperbolic cotangent function by evaluating the limit of the function as $h$ closer to $0$.
$= \,\,\,$ $\dfrac{\coth^{-1}{(x+0)}-\coth^{-1}{x}}{0}$
$= \,\,\,$ $\dfrac{\coth^{-1}{x}-\coth^{-1}{x}}{0}$
$= \,\,\,$ $\require{cancel} \dfrac{\cancel{\coth^{-1}{x}}-\cancel{\coth^{-1}{x}}}{0}$
$=\,\,\,$ $\dfrac{0}{0}$
The derivative of the inverse hyperbolic cotangent function is indeterminate but it is not true because the direct substitution method is failed here.
In this case, we must have to think for an alternative method in order to prove the derivative formula for the inverse hyperbolic cotangent function by the first principle of differentiation.
$\implies$ $\dfrac{d}{dx}{\, (\coth^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\coth^{-1}{(x+h)}-\coth^{-1}{x}}{h}}$
As per the inverse hyperbolic functions, the inverse hyperbolic cotangent function can be written in logarithmic form.
$\coth^{-1}{x}$ $\,=\,$ $\dfrac{1}{2}\,\log_{e}{\Bigg(\dfrac{x+1}{x-1}\Bigg)}$
Then, $\coth^{-1}{(x+h)}$ $\,=\,$ $\dfrac{1}{2} \, \log_{e}{\Bigg(\dfrac{(x+h)+1}{(x+h)-1}\Bigg)}$
Each inverse hyperbolic cot function can be replaced by its equivalent form in log form in the fundamental definition of the differentiation of the inverse hyperbolic cotangent function.
$\implies$ $\dfrac{d}{dx}{\,(\coth^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{1}{2}\,\log_{e}{\Bigg(\dfrac{(x+h)+1}{(x+h)-1}\Bigg)}-\dfrac{1}{2}\,\log_{e}{\Bigg(\dfrac{x+1}{x-1}\Bigg)}}{h} }$
We can now concentrate on simplifying the right hand side mathematical expression of the equation.
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{1}{2}\,\Bigg[\log_{e}{\Bigg(\dfrac{(x+h)+1}{(x+h)-1}\Bigg)}-\log_{e}{\Bigg(\dfrac{x+1}{x-1}\Bigg)}\Bigg]}{h} }$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{1}{2}\,\times\,\dfrac{\log_{e}{\Bigg(\dfrac{(x+h)+1}{(x+h)-1}\Bigg)}-\log_{e}{\Bigg(\dfrac{x+1}{x-1}\Bigg)}}{h}\Bigg]}$
The constant multiple rule of limits is used to separate the constant from the limiting operation.
$=\,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(\dfrac{(x+h)+1}{(x+h)-1}\Bigg)}-\log_{e}{\Bigg(\dfrac{x+1}{x-1}\Bigg)}}{h}}$
Mathematically, expand the logarithmic terms in the numerator by the quotient rule of logarithms.
$=\,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big((x+h)+1\Big)}-\log_{e}{\Big((x+h)-1\Big)}-\Big[\log_{e}{\Big(x+1\Big)}-\log_{e}{\Big(x-1\Big)}\Big]}{h}}$
$=\,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big((x+h)+1\Big)}-\log_{e}{\Big((x+h)-1\Big)}-\log_{e}{\Big(x+1\Big)}+\log_{e}{\Big(x-1\Big)}}{h}}$
$=\,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big((x+h)+1\Big)}-\log_{e}{\Big(x+1\Big)}-\log_{e}{\Big((x+h)-1\Big)}+\log_{e}{\Big(x-1\Big)}}{h}}$
$=\,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big((x+h)+1\Big)}-\log_{e}{\Big(x+1\Big)}-\Big[\log_{e}{\Big((x+h)-1\Big)}-\log_{e}{\Big(x-1\Big)}\Big]}{h}}$
The quotient rule of logarithms can be used one more time for combining the logarithmic terms in the following form.
$=\,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(\dfrac{(x+h)+1}{x+1}\Bigg)}-\Bigg[\log_{e}{\Bigg(\dfrac{(x+h)-1}{x-1}\Bigg)}\Bigg]}{h}}$
$=\,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(\dfrac{(x+h)+1}{x+1}\Bigg)}-\log_{e}{\Bigg(\dfrac{(x+h)-1}{x-1}\Bigg)}}{h}}$
$=\,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(\dfrac{x+h+1}{x+1}\Bigg)}-\log_{e}{\Bigg(\dfrac{x+h-1}{x-1}\Bigg)}}{h}}$
$=\,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(\dfrac{x+1+h}{x+1}\Bigg)}-\log_{e}{\Bigg(\dfrac{x-1+h}{x-1}\Bigg)}}{h}}$
$=\,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(\dfrac{x+1}{x+1}+\dfrac{h}{x+1}\Bigg)}-\log_{e}{\Bigg(\dfrac{x-1}{x-1}+\dfrac{h}{x-1}\Bigg)}}{h}}$
$=\,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(\dfrac{\cancel{x+1}}{\cancel{x+1}}+\dfrac{h}{x+1}\Bigg)}-\log_{e}{\Bigg(\dfrac{\cancel{x-1}}{\cancel{x-1}}+\dfrac{h}{x-1}\Bigg)}}{h}}$
$=\,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{x+1}\Bigg)}-\log_{e}{\Bigg(1+\dfrac{h}{x-1}\Bigg)}}{h}}$
$=\,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{\log_{e}{\Bigg(1+\dfrac{h}{x+1}\Bigg)}}{h}-\dfrac{\log_{e}{\Bigg(1+\dfrac{h}{x-1}\Bigg)}}{h}\Bigg]}$
Use the difference rule of limits to evaluate the limit of difference of the functions.
$=\,\,\,$ $\dfrac{1}{2} \times \Bigg[\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{x+1}\Bigg)}}{h}}$ $-$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{x-1}\Bigg)}}{h}\Bigg]}$
Every term in the expression is almost in the form of logarithmic limit rule but there are some adjustments needed for using the logarithmic limit rule here.
$=\,\,\,$ $\dfrac{1}{2} \times \Bigg[\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{x+1}\Bigg)}}{\dfrac{h \times (x+1)}{x+1}}}$ $-$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{x-1}\Bigg)}}{\dfrac{h \times (x-1)}{x-1}}\Bigg]}$
$=\,\,\,$ $\dfrac{1}{2} \times \Bigg[\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{x+1}\Bigg)}}{\dfrac{h}{x+1} \times (x+1)}}$ $-$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{x-1}\Bigg)}}{\dfrac{h}{x-1} \times (x-1)}\Bigg]}$
$=\,\,\,$ $\dfrac{1}{2} \times \Bigg[\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{x+1}\Bigg)}}{(x+1) \times \dfrac{h}{x+1}}}$ $-$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{x-1}\Bigg)}}{(x-1) \times \dfrac{h}{x-1}}\Bigg]}$
$=\,\,\,$ $\dfrac{1}{2} \times \Bigg[\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{1}{x+1} \times \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{x+1}\Bigg)}}{\dfrac{h}{x+1}}\Bigg)}$ $-$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{1}{x-1} \times \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{x-1}\Bigg)}}{\dfrac{h}{x-1}}\Bigg)\Bigg]}$
In this case, the input of limit operation is in terms of $h$ and the terms in $x$ become constants. So, the functions in $x$ can be separated from the limiting operations. It can be done by the constant multiple rule of limits.
$=\,\,\,$ $\dfrac{1}{2} \times \Bigg[\dfrac{1}{x+1} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{x+1}\Bigg)}}{\dfrac{h}{x+1}}}$ $-$ $\dfrac{1}{x-1} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{x-1}\Bigg)}}{\dfrac{h}{x-1}}\Bigg]}$
$(1)\,\,\,$ Let $h\,\to\,0$ then $\dfrac{h}{x+1}\,\to\,\dfrac{0}{x+1}$. So $\dfrac{h}{x+1}\,\to\,0$. Now, assume $m = \dfrac{h}{x+1}$. Therefore, $m\,\to\,0$.
$(2)\,\,\,$ In similar way, if $h\,\to\,0$ then $\dfrac{h}{x-1}\,\to\,\dfrac{0}{x-1}$. Hence $\dfrac{h}{x-1}\,\to\,0$. Now, take $n = \dfrac{h}{x-1}$. Therefore, $n\,\to\,0$.
As per above two assumptions, the first and second terms in the expression can be expressed in $m$ and $n$ respectively.
$=\,\,\,$ $\dfrac{1}{2} \times \Bigg[\dfrac{1}{x+1} \times \displaystyle \large \lim_{m \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+m\Big)}}{m}}$ $-$ $\dfrac{1}{n-1} \times \displaystyle \large \lim_{n \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+n\Big)}}{n}\Bigg]}$
The limit of every function is one as per the limit rule of logarithms.
$=\,\,\,$ $\dfrac{1}{2} \times \Bigg[\dfrac{1}{x+1} \times 1$ $-$ $\dfrac{1}{x-1} \times 1\Bigg]$
Now, let’s concentrate on simplifying the above mathematical expression by the fundamental operations.
$=\,\,\,$ $\dfrac{1}{2} \times \Bigg[\dfrac{1}{x+1}-\dfrac{1}{x-1}\Bigg]$
$=\,\,\,$ $\dfrac{1}{2} \times \Bigg[\dfrac{1 \times (x-1)-(x+1) \times 1}{(x+1)(x-1)}\Bigg]$
$=\,\,\,$ $\dfrac{1}{2} \times \Bigg[\dfrac{(x-1)-(x+1)}{(x+1)(x-1)}\Bigg]$
$=\,\,\,$ $\dfrac{1}{2} \times \dfrac{(x-1)-(x+1)}{(x+1)(x-1)}$
$=\,\,\,$ $\dfrac{1}{2} \times \dfrac{x-1-x-1}{x^2-1}$
$=\,\,\,$ $\dfrac{1}{2} \times \dfrac{x-x-1-1}{x^2-1}$
$=\,\,\,$ $\dfrac{1}{2} \times \dfrac{\cancel{x}-\cancel{x}-2}{x^2-1}$
$=\,\,\,$ $\dfrac{1}{2} \times \dfrac{-2}{x^2-1}$
$=\,\,\,$ $\dfrac{1 \times (-2)}{2 \times (x^2-1)}$
$=\,\,\,$ $\dfrac{-2}{2 \times (x^2-1)}$
$=\,\,\,$ $\dfrac{-\cancel{2}}{\cancel{2} \times (x^2-1)}$
$=\,\,\,$ $\dfrac{-1}{x^2-1}$
$=\,\,\,$ $\dfrac{1}{1-x^2}$
In this way, we have proved that the differentiation of the inverse hyperbolic cotangent function equals to the multiplicative inverse of the difference of square of variable from one.
$\therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, \coth^{-1}{x}}$ $\,=\,$ $\dfrac{1}{1-x^2}$
Thus, the derivative formula for the inverse hyperbolic co-tangent function can be proved mathematically by the first principle of differentiation in differential calculus.
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