Derivative of Tangent Inverse

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

 

Let the function of the form be

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

 

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

\tan y = x

 

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

\begin{gathered} \frac{d}{{dx}}\tan y = \frac{d}{{dx}}\left( x \right) \\ \Rightarrow {\sec ^2}y\frac{{dy}}{{dx}} = 1 \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{{{\sec }^2}y}}\,\,\,\,{\text{ - - - }}\left( {\text{i}} \right) \\ \end{gathered}

 

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

\begin{gathered} \frac{{dy}}{{dx}} = \frac{1}{{1 + {{\tan }^2}y}} \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{1 + {x^2}}},\,\,\,\,x \in \mathbb{R} \\ \Rightarrow \frac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \frac{1}{{1 + {x^2}}},\,\,\,\,x \in \mathbb{R} \\ \end{gathered}

 

Example: Find the derivative of

y = f\left( x \right) = {\tan ^{ - 1}}\left( {\frac{x}{a}} \right)

We have the given function as

y = {\tan ^{ - 1}}\left( {\frac{x}{a}} \right)

Differentiating with respect to variable x, we get

\frac{{dy}}{{dx}} = \frac{d}{{dx}}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right)

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

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