Derivative of Inverse Hyperbolic Sine

In this tutorial we shall be concerned with the derivative of inverse hyperbolic sine function with an example.

Let the function of the form

y = f\left( x \right) = {\sinh ^{ - 1}}x

By definition of inverse trigonometric function, y = {\sinh ^{ - 1}}x can be written as

\sinh  y = x

Differentiating both sides with respect to the variable x, we have

\begin{gathered} \frac{d}{{dx}}\sinh y = \frac{d}{{dx}}\left(  x \right) \\ \Rightarrow \cosh y\frac{{dy}}{{dx}} = 1 \\ \Rightarrow \frac{{dy}}{{dx}} =  \frac{1}{{\cosh y}}\,\,\,\,{\text{ - - - }}\left( {\text{i}} \right) \\ \end{gathered}


We can write from the fundamental rules of inverse hyperbolic identities \cosh y = \sqrt {1 + {{\sinh }^2}y} . Putting this value in above relation (i) and simplifying, we have

\frac{{dy}}{{dx}}  = \frac{1}{{\sqrt {1 + {{\sinh }^2}y} }}


From the above we have \sinh  y = x, by putting this

\begin{gathered} \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 +  {x^2}} }} \\ \Rightarrow \frac{d}{{dx}}\left( {{{\sinh  }^{ - 1}}x} \right) = \frac{1}{{\sqrt {1 + {x^2}} }} \\ \end{gathered}

Example: Find the derivative of

y = f\left( x \right)  = {\sinh ^{ - 1}}4x

We have the given function as

y =  {\sinh ^{ - 1}}4x

Differentiation with respect to variable x, we get

\frac{{dy}}{{dx}}  = \frac{d}{{dx}}{\sinh ^{ - 1}}4x


Using the rule, \frac{d}{{dx}}\left(  {{{\sinh }^{ - 1}}x} \right) = \frac{1}{{\sqrt {1 + {x^2}} }}, we get

\begin{gathered} \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 +  {{\left( {4x} \right)}^2}} }}\frac{d}{{dx}}4x \\ \Rightarrow \frac{{dy}}{{dx}} =  \frac{4}{{\sqrt {1 + 16{x^2}} }} \\ \end{gathered}

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