Derivative of Inverse Hyperbolic Cosecant
In this tutorial we shall discuss the derivative of the inverse hyperbolic cosecant function with an example.
Let the function be of the form \[y = f\left( x \right) = {\operatorname{csch} ^{ – 1}}x\]
By the definition of the inverse trigonometric function, $$y = {\operatorname{csch} ^{ – 1}}x$$ can be written as
\[\operatorname{csch} y = x\]
Differentiating both sides with respect to the variable $$x$$, we have
\[\begin{gathered} \frac{d}{{dx}}\operatorname{csch} y = \frac{d}{{dx}}\left( x \right) \\ \Rightarrow – \csc {\text{h}}y\coth y\frac{{dy}}{{dx}} = 1 \\ \Rightarrow \frac{{dy}}{{dx}} = – \frac{1}{{\csc {\text{h}}y\coth y}}\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ \end{gathered} \]
From the fundamental rules of inverse hyperbolic identities, this can be written as $$\coth y = \sqrt {1 + \csc {{\text{h}}^2}x} $$. Putting this value in above relation (i) and simplifying, we have
\[\frac{{dy}}{{dx}} = – \frac{1}{{\csc {\text{h}}x\sqrt {1 + \csc {{\text{h}}^2}x} }}\]
From the above we have $$\operatorname{csch} y = x$$, thus
\[\begin{gathered} \frac{{dy}}{{dx}} = – \frac{1}{{x\sqrt {1 + {x^2}} }} \\ \Rightarrow \frac{d}{{dx}}\left( {{{\operatorname{csch} }^{ – 1}}x} \right) = – \frac{1}{{x\sqrt {1 + {x^2}} }} \\ \end{gathered} \]
Example: Find the derivative of \[y = f\left( x \right) = {\operatorname{csch} ^{ – 1}}2x\]
We have the given function as
\[y = {\operatorname{csch} ^{ – 1}}2x\]
Differentiating with respect to variable $$x$$, we get
\[\frac{{dy}}{{dx}} = \frac{d}{{dx}}{\operatorname{csch} ^{ – 1}}2x\]
Using the rule, $$\frac{d}{{dx}}\left( {{{\operatorname{csch} }^{ – 1}}x} \right) = – \frac{1}{{x\sqrt {1 + {x^2}} }}$$, we get
\[\begin{gathered} \frac{{dy}}{{dx}} = – \frac{1}{{2x\sqrt {1 – {{\left( {2x} \right)}^2}} }}\frac{d}{{dx}}2x \\ \Rightarrow \frac{{dy}}{{dx}} = – \frac{1}{{2x\sqrt {1 – 4{x^2}} }}2\left( 1 \right) \\ \Rightarrow \frac{{dy}}{{dx}} = – \frac{1}{{x\sqrt {1 – 4{x^2}} }} \\ \end{gathered} \]
Greg Hullender
May 31 @ 12:17 am
I’m pretty sure you’re missing an |x| in the answer. That’s because when you converted coth to sqrt(1+csch^2) you forgot that coth can be negative in the range of csch. But because coth and csch have the same sign, all you have to do to fix it is add the absolute value.