Derivative of Secant Inverse

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

Let the function be of the form

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

By the definition of the 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}

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

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

Now we have \sec y = x, and putting this value in the 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

Differentiating 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}