Derivative of Secant Inverse

In this tutorial we shall explore the derivative of inverse trigonometric functions and we shall prove derivative of secant inverse.

Let the function of the form

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

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

\sec  y = x


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

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


We can write from the fundamental trigonometric rules \tan y =  \sqrt {{{\sec }^2}y - 1} . Putting this value in above relation (i) and simplifying, we have

\frac{{dy}}{{dx}}  = \frac{1}{{\sec y\sqrt {{{\sec }^2}y - 1} }}


But we have \sec y = x, putting this value in above relation

\begin{gathered} \frac{{dy}}{{dx}} = \frac{1}{{x\sqrt {{x^2} -  1} }},\,\,\,\,x \in \mathbb{R} - \left[ { - 1,1} \right] \\ \Rightarrow \frac{d}{{dx}}\left( {{{\sec }^{  - 1}}x} \right) = \frac{1}{{x\sqrt {{x^2} - 1} }},\,\,\,\,x \in \mathbb{R} -  \left[ { - 1,1} \right] \\ \end{gathered}

Example: Find the derivative of

y = f\left( x \right)  = {\sec ^{ - 1}}2x

We have the given function as

y =  {\sec ^{ - 1}}2x


Differentiation with respect to variable x, we get

\frac{{dy}}{{dx}}  = \frac{d}{{dx}}{\sec ^{ - 1}}2x


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

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

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