Derivative Of Inverse Hyperbolic Sine | EMathZone

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

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

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

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

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

Example: Find the derivative of \[y = f\left( x \right) = {\sinh ^{ – 1}}4x\]

We have the given function as \[y = {\sinh ^{ – 1}}4x\]

Differentiating 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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