Derivative of Inverse Hyperbolic Tangent function with respect to x

Formula

$\dfrac{d}{dx} \, \tanh^{-1} x \,=\, \dfrac{1}{1-x^2}$

Proof

$x$ is a literal number and represents a real number. The inverse hyperbolic tangent function is written as $\tanh^{-1} x$ in mathematics and the derivative of inverse hyperbolic tangent function with respect to $x$ is written as follows.

$\dfrac{d}{dx} \, \tanh^{-1} x$

01

Differentiation in Limit form

The differentiation of a function can be expressed in limit form and this rule is used to prove the derivative of $\tanh^{-1}{x}$ with respect to $x$.

$\dfrac{d}{dx} \, f(x)$ $\,=\,$ $\displaystyle \large \lim_{h \to 0}$ $\dfrac{f(x+h)-f(x)}{h}$

Take, $f(x) \,=\, \tanh^{-1}{x}$ then $f(x+h) \,=\, \tanh^{-1}{(x+h)}$

Now, express the derivative of inverse hyperbolic tangent function with respect to $x$ in limit form.

$\dfrac{d}{dx} \, \tanh^{-1}{x}$ $\,=\,$ $\displaystyle \large \lim_{h \to 0}$ $\dfrac{\tanh^{-1}{(x+h)}-\tanh^{-1}{x}}{h}$

02

Transforming Inverse Hyperbolic Tangent function

The inverse hyperbolic tangent function can be expressed in logarithmic form.

$\tanh^{-1}{x} \,=\, \dfrac{1}{2} \, \log_{e}{\Bigg(\dfrac{1+x}{1-x}\Bigg)}$

Therefore, $\tanh^{-1}{(x+h)} \,=\, \dfrac{1}{2} \, \log_{e}{\Bigg(\dfrac{1+(x+h)}{1-(x+h)}\Bigg)}$

Replace the inverse hyperbolic tangent functions by its equivalent logarithmic form expressions.

$\implies \dfrac{d}{dx} \, \tanh^{-1}{x}$ $\,=\,$ $\displaystyle \large \lim_{h \to 0}$ $\dfrac{ \dfrac{1}{2} \, \log_{e}{\Bigg(\dfrac{1+(x+h)}{1-(x+h)}\Bigg)}-\dfrac{1}{2} \, \log_{e}{\Bigg(\dfrac{1+x}{1-x}\Bigg)}}{h}$

Take the half common from both terms in the numerator of the fraction.

$\implies \dfrac{d}{dx} \, \tanh^{-1}{x}$ $\,=\,$ $\displaystyle \large \lim_{h \to 0}$ $\dfrac{ \dfrac{1}{2} \Bigg[\log_{e}{\Bigg(\dfrac{1+(x+h)}{1-(x+h)}\Bigg)} \,-\, \log_{e}{\Bigg(\dfrac{1+x}{1-x}\Bigg)\Bigg]}}{h}$

$=\,$ $\displaystyle \large \lim_{h \to 0}$ $\dfrac{\log_{e}{\Bigg(\dfrac{1+(x+h)}{1-(x+h)}\Bigg)} \,-\, \log_{e}{\Bigg(\dfrac{1+x}{1-x}\Bigg)}}{2h}$

Apply quotient rule of logarithms to expand the log of quotient of two terms as subtraction of logs of them.

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \dfrac{\log_{e}{[1+(x+h)]}-\log_{e}{[1-(x+h)]} \,-\, [\log_{e}{(1+x)}-\log_{e}{(1-x)} ]}{2h}$

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \dfrac{\log_{e}{[1+(x+h)]}-\log_{e}{[1-(x+h)]}-\log_{e}{(1+x)}+\log_{e}{(1-x)}}{2h}$

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \dfrac{\log_{e}{[1+(x+h)]}-\log_{e}{(1+x)}-\log_{e}{[1-(x+h)]}+\log_{e}{(1-x)}}{2h}$

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \dfrac{\log_{e}{[1+(x+h)]}-\log_{e}{(1+x)}-[\log_{e}{[1-(x+h)]}-\log_{e}{(1-x)}]}{2h}$

Once again use quotient rule of logarithms to express subtraction of logs of terms in logs of quotient of them.

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \dfrac{\log_{e}{\Bigg( \dfrac{1+(x+h)}{1+x} \Bigg)}-\log_{e}{\Bigg(\dfrac{1-(x+h)}{1-x} \Bigg)}}{2h}$

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \dfrac{\log_{e}{\Bigg( \dfrac{(1+x)+h}{1+x} \Bigg)}-\log_{e}{\Bigg(\dfrac{(1-x)-h)}{1-x} \Bigg)}}{2h}$

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \dfrac{\log_{e}{\Bigg(\dfrac{(1+x)}{1+x}+\dfrac{h}{1+x}\Bigg)}-\log_{e}{\Bigg(\dfrac{(1-x)}{1-x}-\dfrac{h}{1-x} \Bigg)}}{2h}$

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \require{cancel} \dfrac{\log_{e}{\Bigg(\dfrac{\cancel{1+x}}{\cancel{1+x}}+\dfrac{h}{1+x}\Bigg)}-\log_{e}{\Bigg(\dfrac{\cancel{1-x}}{\cancel{1-x}}-\dfrac{h}{1-x} \Bigg)}}{2h}$

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{1+x}\Bigg)}-\log_{e}{\Bigg(1-\dfrac{h}{1-x}\Bigg)}}{2h}$

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{1+x}\Bigg)}-\log_{e}{\Bigg(1-\dfrac{h}{1-x}\Bigg)}}{2h}$

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \Bigg[ \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{1+x}\Bigg)}}{2h}-\dfrac{\log_{e}{\Bigg(1-\dfrac{h}{1-x}\Bigg)}}{2h} \Bigg]$

The limit belongs to both terms. So, it can be applied to each term.

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{1+x}\Bigg)}}{2h}$ $\,-\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \dfrac{\log_{e}{\Bigg(1-\dfrac{h}{1-x}\Bigg)}}{2h}$

Limit of the each function is required to find as the value of $h$ tends to $0$. So, evaluate them and replace their values in this expression and it is done in the next step.

03

Expanding Logarithmic functions

The logarithmic function can be expressed as Maclaurin series.

$\log_{e}{(1+x)} \,=\, x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots$

According to the expansion of the logarithmic function as Maclaurin series.

$\displaystyle \large \lim_{h \to 0} \normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{1+x}\Bigg)}}{2h}$ $\,=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \Bigg(\dfrac{\dfrac{h}{1+x}-\dfrac{{\Bigg[\dfrac{h}{1+x}\Bigg]}^2}{2}+\dfrac{{\Bigg[\dfrac{h}{1+x}\Bigg]}^3}{3}-\cdots}{2h}\Bigg)$

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \, \dfrac{1}{2h} \Bigg(\dfrac{h}{1+x}-\dfrac{{\Bigg[\dfrac{h}{1+x}\Bigg]}^2}{2}+\dfrac{{\Bigg[\dfrac{h}{1+x}\Bigg]}^3}{3}-\cdots\Bigg)$

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \, \dfrac{1}{2h} \Bigg(\dfrac{h}{1+x}-\dfrac{h^2}{2{(1+x)}^2}+\dfrac{h^3}{3{(1+x)}^3}-\cdots\Bigg)$

Take $h$ common from all the terms.

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \, \dfrac{h}{2h} \Bigg(\dfrac{1}{1+x}-\dfrac{h}{2{(1+x)}^2}+\dfrac{h^2}{3{(1+x)}^3}-\cdots\Bigg)$

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \, \require{cancel} \dfrac{\cancel{h}}{2\cancel{h}} \Bigg(\dfrac{1}{1+x}-\dfrac{h}{2{(1+x)}^2}+\dfrac{h^2}{3{(1+x)}^3}-\cdots\Bigg)$

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \, \dfrac{1}{2} \Bigg(\dfrac{1}{1+x}-\dfrac{h}{2{(1+x)}^2}+\dfrac{h^2}{3{(1+x)}^3}-\cdots\Bigg)$

Now, substitute $h = 0$ and find its value.

$=\,$ $\dfrac{1}{2} \Bigg(\dfrac{1}{1+x}-\dfrac{0}{2{(1+x)}^2}+\dfrac{{(0)}^2}{3{(1+x)}^3}-\cdots\Bigg)$

$=\,$ $\dfrac{1}{2} \Bigg(\dfrac{1}{1+x}-0+0-\cdots\Bigg)$

$=\,$ $\dfrac{1}{2} \times \dfrac{1}{1+x}$

$\therefore \,\,\, \displaystyle \large \lim_{h \to 0} \normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{1+x}\Bigg)}}{2h}$ $\,=\,$ $\dfrac{1}{2(1+x)}$

Similarly, the logarithmic function can be expressed as Maclaurin series.

$\log_{e}{(1-x)} \,=\, -\Big[x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\cdots\Big]$

According to the expansion of the logarithmic function as Maclaurin series.

$\displaystyle \large \lim_{h \to 0} \normalsize \dfrac{\log_{e}{\Bigg(1-\dfrac{h}{1-x}\Bigg)}}{2h}$ $\,=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize -\Bigg(\dfrac{\dfrac{h}{1-x}+\dfrac{{\Bigg[\dfrac{h}{1-x}\Bigg]}^2}{2}+\dfrac{{\Bigg[\dfrac{h}{1-x}\Bigg]}^3}{3}+\cdots}{2h}\Bigg)$

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \, -\dfrac{1}{2h} \Bigg(\dfrac{h}{1-x}+\dfrac{{\Bigg[\dfrac{h}{1-x}\Bigg]}^2}{2}+\dfrac{{\Bigg[\dfrac{h}{1-x}\Bigg]}^3}{3}+\cdots\Bigg)$

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \, -\dfrac{1}{2h} \Bigg(\dfrac{h}{1-x}+\dfrac{h^2}{2{(1-x)}^2}+\dfrac{h^3}{3{(1-x)}^3}+\cdots\Bigg)$

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \, -\dfrac{h}{2h} \Bigg(\dfrac{1}{1-x}+\dfrac{h}{2{(1-x)}^2}+\dfrac{h^2}{3{(1-x)}^3}+\cdots\Bigg)$

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \, \require{cancel} -\dfrac{\cancel{h}}{2\cancel{h}} \Bigg(\dfrac{1}{1-x}+\dfrac{h}{2{(1-x)}^2}+\dfrac{h^2}{3{(1-x)}^3}+\cdots\Bigg)$

$=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \, -\dfrac{1}{2} \Bigg(\dfrac{1}{1-x}+\dfrac{h}{2{(1-x)}^2}+\dfrac{h^2}{3{(1-x)}^3}+\cdots\Bigg)$

$=\,$ $-\dfrac{1}{2} \Bigg(\dfrac{1}{1-x}-\dfrac{0}{2{(1-x)}^2}+\dfrac{{(0)}^2}{3{(1-x)}^3}-\cdots\Bigg)$

$=\,$ $-\dfrac{1}{2} \Bigg(\dfrac{1}{1-x}+0+0+\cdots\Bigg)$

$=\,$ $-\dfrac{1}{2} \times \dfrac{1}{1-x}$

$\therefore \,\,\, \displaystyle \large \lim_{h \to 0} \normalsize \dfrac{\log_{e}{\Bigg(1-\dfrac{h}{1-x}\Bigg)}}{2h}$ $\,=\,$ $-\dfrac{1}{2(1-x)}$

04

Simplifying the functions

Comeback to actual step of the simplification of the differentiation of inverse hyperbolic tangent function.

$\implies \dfrac{d}{dx} \, \tanh^{-1} x$ $\,=\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{1+x}\Bigg)}}{2h}$ $\,-\,$ $\displaystyle \large \lim_{h \to 0} \normalsize \dfrac{\log_{e}{\Bigg(1-\dfrac{h}{1-x}\Bigg)}}{2h}$

Now, replace each limit of the function as $h$ approaches to zero by its respective value.

$\implies \dfrac{d}{dx} \, \tanh^{-1} x$ $\,=\,$ $\dfrac{1}{2(1+x)}$ $\,-\,$ $\Bigg[-\dfrac{1}{2(1-x)}\Bigg]$

$\implies \dfrac{d}{dx} \, \tanh^{-1} x$ $\,=\,$ $\dfrac{1}{2(1+x)}+\dfrac{1}{2(1-x)}$

$\implies \dfrac{d}{dx} \, \tanh^{-1} x$ $\,=\,$ $\dfrac{1}{2} \Bigg[\dfrac{1}{1+x}+\dfrac{1}{1-x}\Bigg]$

$\implies \dfrac{d}{dx} \, \tanh^{-1} x$ $\,=\,$ $\dfrac{1}{2} \Bigg[\dfrac{1-x+1+x}{(1+x)(1-x)}\Bigg]$

$\implies \dfrac{d}{dx} \, \tanh^{-1} x$ $\,=\,$ $\require{cancel} \dfrac{1}{2} \Bigg[\dfrac{1-\cancel{x}+1+\cancel{x}}{1-x^2}\Bigg]$

$\implies \dfrac{d}{dx} \, \tanh^{-1} x$ $\,=\,$ $\dfrac{2}{2(1-x^2)}$

$\implies \dfrac{d}{dx} \, \tanh^{-1} x$ $\,=\,$ $\require{cancel} \dfrac{\cancel{2}}{\cancel{2}(1-x^2)}$

$\,\,\, \therefore \,\,\,\,\,\, \dfrac{d}{dx} \, \tanh^{-1} x$ $\,=\,$ $\dfrac{1}{1-x^2}$

Therefore, it is proved that the derivative of inverse hyperbolic tangent function with respect to $x$ is equal to one divided by $1$ minus $x$ squared. The differentiation property is used as a law in differentiating the inverse hyperbolic functions in the differentiation.