# Proof of Derivative of Inverse Hyperbolic Tangent function

The inverse hyperbolic tangent in function form is written as $\tanh^{-1}{x}$ or $\operatorname{arctanh}{x}$ when $x$ is considered to represent a variable. The differentiation or the derivative of inverse hyperbolic tan function with respect to $x$ is written in two different mathematical forms as follows.

$(1).\,\,\,$ $\dfrac{d}{dx}{\, \Big(\tanh^{-1}{(x)}\Big)}$

$(2).\,\,\,$ $\dfrac{d}{dx}{\, \Big(\operatorname{arctanh}{(x)}\Big)}$

The derivative rule of inverse hyperbolic tangent function is proved mathematically from the first principle of differentiation in calculus. Now, let us study how to prove the derivative rule for inverse hyperbolic tan function in mathematics.

### Derivative of Inverse Hyperbolic Tan in Limit form

The differentiation of the inverse hyperbolic tangent function can be derived in limit form by the fundamental definition of the derivative.

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

When the differential element $\Delta x$ is simply represented by $h$ for our convenience, then the whole mathematical expression can be expressed in terms of $h$ instead of $\Delta x$.

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

Mathematically, the derivative rule of $\operatorname{arctanh}{(x)}$ function with respect to $x$ is actually derived in differential calculus from the first principle of differentiation.

### Evaluate the Limit by the Direct Substitution

In limits, the direct substitution method is a basic method for evaluating the limit of any mathematical expression. So, let us try to derive the differentiation of inverse hyperbolic tangent function by evaluating the limit of the mathematical expression as $h$ approaches $0$.

$= \,\,\,$ $\dfrac{\tanh^{-1}{(x+0)}-\tanh^{-1}{x}}{0}$

$= \,\,\,$ $\dfrac{\tanh^{-1}{x}-\tanh^{-1}{x}}{0}$

$= \,\,\,$ $\require{cancel} \dfrac{\cancel{\tanh^{-1}{x}}-\cancel{\tanh^{-1}{x}}}{0}$

$=\,\,\,$ $\dfrac{0}{0}$

It is evaluated that the derivative of the inverse hyperbolic tangent function is indeterminate. Actually, it is not correct due to the failure of the direct substitution method.

### Simplify the mathematical expression

Now, we have to think for an alternative approach for proving the differentiation formula for the inverse hyperbolic tangent function from first principle of differentiation.

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

According to the Inverse hyperbolic functions, the inverse hyperbolic tangent function can be expressed in logarithmic form.

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

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

Now, replace every inverse hyperbolic tangent function by their equivalent logarithmic form expressions in the fundamental definition of the derivative of inverse hyperbolic tan function.

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

Now, let us focus on simplifying the mathematical expression in the right hand side of the equation.

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

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

The constant be separated by the constant multiple rule of limits.

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

Every logarithmic term in the numerator can be expanded by the quotient rule of logarithms.

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

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

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

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

Use the same quotient rule of logarithms to combine the logarithmic terms as follows.

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

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

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

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

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

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

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

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

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

Now, we can use the addition rule of limits for evaluating the limit of sum of the functions.

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

Each term in the expression is in the form of logarithmic limit rule but it requires some adjustments.

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

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

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

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

The input of the limiting operation is in terms of $h$. So, the terms in $x$ are constants and they can be excluded from the limit operation. Hence, use the constant multiple rule of limits to separate the constants from limiting operation.

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

### Evaluate the Limit of the functions

$(1)\,\,\,$ When $h\,\to\,0$ then $\dfrac{h}{1+x}\,\to\,\dfrac{0}{1+x}$. Therefore $\dfrac{h}{1+x}\,\to\,0$. Now, take $y = \dfrac{h}{1+x}$. Hence, $y\,\to\,0$.

$(2)\,\,\,$ Similarly, when $h\,\to\,0$ then $-h\,\to\,0$, and then $\dfrac{-h}{1-x}\,\to\,\dfrac{0}{1-x}$. Therefore $\dfrac{-h}{1-x}\,\to\,0$. Now, take $z = \dfrac{-h}{1-x}$. Hence, $z\,\to\,0$.

Now, we can express the first term in the expression in terms of $y$ and second term in $z$ by our assumption.

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

The limit of each function is one as per the logarithmic limit rule.

$=\,\,\,$ $\dfrac{1}{2} \times \Bigg[\dfrac{1}{1+x} \times 1$ $+$ $\dfrac{1}{1-x} \times 1\Bigg]$

Now, simplify the whole expression by the mathematical operation.

$=\,\,\,$ $\dfrac{1}{2} \times \Bigg[\dfrac{1}{1+x}+\dfrac{1}{1-x}\Bigg]$

$=\,\,\,$ $\dfrac{1}{2} \times \Bigg[\dfrac{1 \times (1-x)+(1+x) \times 1}{(1+x)(1-x)}\Bigg]$

$=\,\,\,$ $\dfrac{1}{2} \times \Bigg[\dfrac{(1-x)+(1+x)}{(1+x)(1-x)}\Bigg]$

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

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

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

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

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

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

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

$=\,\,\,$ $\dfrac{\cancel{2}}{\cancel{2} \times (1-x^2)}$

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

Therefore, we have proved that the differentiation of the inverse hyperbolic tangent function equals to the reciprocal of the difference of square of variable from one.

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

In this way, the derivative rule of the inverse hyperbolic tangent function is derived mathematically in differential calculus by the first principle of differentiation.

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