Question

What is the derivative of $f\left( x \right)={{\tan }^{-1}}x$ ?
(a) 1 + x
(b) $\dfrac{1}{1+x}$
(c) $\dfrac{1}{1+{{x}^{2}}}$
(d) None of these

Answer 1

In the given problem, we are trying to find the derivative of the given function in the problem. The function is an inverse trigonometric function and thus we are to find its derivative in a different way. We start by considering $y={{\tan }^{-1}}x$ and then differentiate both sides with respect to x. Once we have done that, we can put back the value of tan y to reach the solution in the form of x and get our desired result.

Complete step by step solution:
According to the problem, we are trying to find the derivative of $f\left( x \right)={{\tan }^{-1}}x$.
So, we start by considering, $y=f\left( x \right)$
Thus, now, we can write, $y={{\tan }^{-1}}x$
This also implies, $x=\tan y$
Now, if we differentiate them, we will reach nearer to our result.
So, differentiating both sides with respect to x, we get,
$\Rightarrow \dfrac{d}{dx}\left( x \right)=\dfrac{d}{dx}\left( \tan y \right)$
Now, $\dfrac{d}{dx}\left( x \right)$gives us the value 1. Writing the value in the equation,
$\Rightarrow 1=\dfrac{d}{dx}\left( \tan y \right)\times \dfrac{dy}{dy}$
As, there is only a term with y, we are using this property.
It can also be written as,
$\Rightarrow 1=\dfrac{d}{dy}\left( \tan y \right)\times \dfrac{dy}{dx}$
Again, $\dfrac{d}{dy}\left( \tan y \right)$gives us the value ${{\sec }^{2}}y$ .
So, we have,
$\Rightarrow 1={{\sec }^{2}}y\times \dfrac{dy}{dx}$
From, ${{\sec }^{2}}x-{{\tan }^{2}}x=1$ , we also have, ${{\sec }^{2}}y={{\tan }^{2}}y+1$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{1+{{\tan }^{2}}y}$
Again, we also have, $y={{\tan }^{-1}}x$,
This also implies, $x=\tan y$
So, we will have the value of ${{\tan }^{2}}y$ as ${{x}^{2}}$ .
Putting the value into equation we got,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}$
Hence, from this, it can be concluded that, $\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}}$.

So, the correct answer is “Option c”.

Note: We have dealt with the trigonometric inverse functions while solving this problem. The inverse functions exist when appropriate restrictions are placed on the domain of the original functions. The domains of the other trigonometric functions are restricted appropriately, so that they become one-to-one functions and their inverse can be determined.