Derivative Of Inverse Hyperbolic Tangent - EMathZone

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In this tutorial we shall discuss the derivative of the inverse hyperbolic tangent function with an example.

Let the function be of the form \[y = f\left( x \right) = {\tanh ^{ – 1}}x\]

By the definition of the inverse trigonometric function, $$y = {\tanh ^{ – 1}}x$$ can be written as \[\tanh y = x\]

Differentiating both sides with respect to the variable $$x$$, we have \[\begin{gathered} \frac{d}{{dx}}\tanh y = \frac{d}{{dx}}\left( x \right) \\ \Rightarrow {\operatorname{sech} ^2}y\frac{{dy}}{{dx}} = 1 \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{{{\operatorname{sech} }^2}y}}\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ \end{gathered} \]

From the fundamental rules of inverse hyperbolic identities, this can be written as  $${\operatorname{sech} ^2}y = 1 – {\tanh ^2}y$$. Putting this value in the above relation (i) and simplifying, we have \[\frac{{dy}}{{dx}} = \frac{1}{{1 – {{\tanh }^2}y}}\]

From the above we have $$\tanh y = x$$, thus \[\begin{gathered} \frac{{dy}}{{dx}} = \frac{1}{{1 – {x^2}}} \\ \Rightarrow \frac{d}{{dx}}\left( {{{\tanh }^{ – 1}}x} \right) = \frac{1}{{1 – {x^2}}} \\ \end{gathered} \]

Example: Find the derivative of \[y = f\left( x \right) = {\tanh ^{ – 1}}{x^2}\]

We have the given function as \[y = {\tanh ^{ – 1}}{x^2}\]

Differentiating with respect to variable $$x$$, we get \[\frac{{dy}}{{dx}} = \frac{d}{{dx}}{\tanh ^{ – 1}}{x^2}\]

Using the rule, $$\frac{d}{{dx}}\left( {{{\tanh }^{ – 1}}x} \right) = \frac{1}{{1 – {x^2}}}$$, we get \[\begin{gathered} \frac{{dy}}{{dx}} = \frac{1}{{1 – {{\left( {{x^2}} \right)}^2}}}\frac{d}{{dx}}{x^2} \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{1 – {x^4}}}2x \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{2x}}{{1 – {x^4}}} \\ \end{gathered} \]

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