Derivative of Tangent Inverse

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

Let the function of the form

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

By definition of 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}


We can write from the fundamental trigonometric rules 1 + {\tan  ^2}y = {\sec ^2}y. Putting this value in 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)


Differentiation 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}

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