Question

What is the derivative of $f\left( x \right) = {\tan ^{ - 1}}\left( x \right)$ ?

Answer 1

We have to deal with an inverse trigonometric function in this question. As the name suggests these are inverse trigonometric functions with restricted domain. The derivatives for standard anti-trigonometric functions should be kept in mind to tackle this kind of questions easily. Some of the derivatives are given as; $\left( 1 \right)\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }}$ . $\left( 2 \right)\dfrac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right) = - \dfrac{1}{{\sqrt {1 - {x^2}} }}$ . $\left( 3 \right)\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \dfrac{1}{{1 + {x^2}}}$ . $\left( 4 \right)\dfrac{d}{{dx}}\left( {{{\cot }^{ - 1}}x} \right) = - \dfrac{1}{{1 + {x^2}}}$ . $\left( 5 \right)\dfrac{d}{{dx}}\left( {{{\sec }^{ - 1}}x} \right) = \dfrac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}$ . $\left( 6 \right)\dfrac{d}{{dx}}\left( {{{\operatorname{cosec} }^{ - 1}}x} \right) = - \dfrac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}$ .

Complete step by step answer:
Let $y = {\tan ^{ - 1}}x{\text{ }}......\left( 1 \right)$
Multiplying by tangent function on both sides, we get;
$\Rightarrow \tan y = \tan \left( {{{\tan }^{ - 1}}x} \right)$
We know that tangent function and its inverse function cancels each other, therefore we get;
$\Rightarrow x = \tan y{\text{ }}......\left( 2 \right)$
Differentiating equation w.r.t. $x$ ;
$\Rightarrow 1 = {\sec ^2}y \times \dfrac{{dy}}{{dx}}{\text{ }}\left( {\because \dfrac{d}{{dx}}\left( {\tan x} \right) = {{\sec }^2}x} \right)$
Rearranging the above equation, we get;
$\Rightarrow \dfrac{1}{{{{\sec }^2}y}} = \dfrac{{dy}}{{dx}}$
We know the relation between tangent function and secant function;
$\left( {\because {{\sec }^2}x - {{\tan }^2}x = 1} \right)$
$\therefore {\sec ^2}x = 1 + {\tan ^2}x$
Therefore, replacing the value of ${\sec ^2}y$ from the above property;
$\Rightarrow \dfrac{1}{{1 + {{\tan }^2}y}} = \dfrac{{dy}}{{dx}}$
From equation $\left( 2 \right)$ , we know that $x = \tan y$ ;
$\Rightarrow \dfrac{1}{{1 + {x^2}}} = \dfrac{{dy}}{{dx}}$
And also if $x = \tan y$, then $y = {\tan ^{ - 1}}x$ ;
So, finally we get; $\dfrac{1}{{1 + {x^2}}} = \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right)$
Therefore the derivative of ${\tan ^{ - 1}}x$ is $\dfrac{1}{{1 + {x^2}}}$ .
We should also be familiar with the concept of domain and range here. The domain of a function represents the complete set of possible values that the independent variable can take and for which the function can exist. While the range represents the set of output values of the dependent variable after putting the values of the domain. Therefore the domain and range of a function are like possible inputs and outputs for a given function.
Note: Trigonometric and inverse trigonometric functions are differentiable only in their respective domain. Let us see some properties of inverse trigonometric functions here. $\left( 1 \right)$ The function ${\sin ^{ - 1}}x$ should not be confused with ${\left( {\sin x} \right)^{ - 1}}$ , since ${\left( {\sin x} \right)^{ - 1}} = \dfrac{1}{{\sin x}}$ and this is true for all other trigonometric identities as well. $\left( 2 \right)$ Some of the properties of inverse trigonometric functions are stated as: $\left( a \right){\sin ^{ - 1}}\dfrac{1}{x} = \cos e{c^{ - 1}}x$ . $\left( b \right){\cos ^{ - 1}}\dfrac{1}{x} = {\sec ^{ - 1}}x$ . $\left( c \right){\tan ^{ - 1}}\dfrac{1}{x} = {\cot ^{ - 1}}x$ . $\left( d \right){\sin ^{ - 1}}\left( { - x} \right) = - {\sin ^{ - 1}}x$ (since sine is an odd function). $\left( e \right){\cos ^{ - 1}}\left( { - x} \right) = \pi - {\cos ^{ - 1}}x$ . $\left( d \right){\tan ^{ - 1}}\left( { - x} \right) = - {\tan ^{ - 1}}x$ .