# 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

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

Harry Dunleavy

June 20@ 8:37 amVery nicely done with enough details. Plenty of details help. I’m also a math teacher.

Dale J

January 14@ 8:56 pmSome textbooks have absolute value around the x in the denominator, but I can’t really find an explanation why! Anyone have an idea?

Marvin Littman

February 23@ 12:22 amHello Dale J,

The reason for the absolute value: In the proof above, a square root was taken, so there should really be a plus or minus in front of the (positive) square root. Also, x itself may be negative. But the slope of the graph of the inverse secant is always positive, which means the derivative of arcsec must be positive. Hence, the absolute value is included to ensure that the derivative is positive.