Question

The derivative of $f\left( {{\tan }^{-1}}x \right),$ where $f\left( x \right)=\tan x$ is
$\left( a \right)\,1$
$\left( b \right)\,\dfrac{1}{1+{{x}^{2}}}$
$\left( c \right)\,2$
$\left( d \right)\,\dfrac{-1}{1+{{x}^{2}}}$

Answer 1

Hint: We will use some familiar trigonometric identity to solve this question. We will substitute for $x$ in the given definition of the function $f.$ Then, we will use the trigonometric identity given by $\tan \left( {{\tan }^{-1}}x \right)=x.$ Then, we will find the derivative.

Complete step by step solution:
Let us consider the given problem.
We are asked to find the derivative of the function $f\left( {{\tan }^{-1}}x \right).$
We know that the function $f$ is defined as $f\left( x \right)=\tan x.$
So, we know that we need to substitute the value inside the brackets on the left-hand side for $x$ on the right-hand side.
So, $f\left( x \right)$ will become $f\left( {{\tan }^{-1}}x \right)$ and so the right-hand side will become $\tan \left( {{\tan }^{-1}}x \right).$
Thus, we will get the function as $f\left( {{\tan }^{-1}}x \right)=\tan \left( {{\tan }^{-1}}x \right).$
We know the trigonometric identity given by $\tan \left( {{\tan }^{-1}}x \right)=x.$
So, when we substitute this in the definition of the function $f,$ we will get $f\left( {{\tan }^{-1}}x \right)=x.$
So, we have simplified the given function.
Now, we need to find the derivative of the function.
And we will get $\dfrac{d}{dx}f\left( {{\tan }^{-1}}x \right)=\dfrac{dx}{dx}.$
We know that $\dfrac{dx}{dx}=1.$
So, we will get the derivative of the function as $\dfrac{d}{dx}f\left( {{\tan }^{-1}}x \right)=\dfrac{dx}{dx}=1.$

Hence the derivative of the function $f\left( {{\tan }^{-1}}x \right)$ is equal to $1$ where $f\left( x \right)=\tan x.$

Note:
We should always remember the trigonometric identities given by $\cos \left( {{\cos }^{-1}}x \right)=x$ and $\sin \left( {{\sin }^{-1}}x \right)=x.$ We know the basic rule of differentiation which is given by $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}.$ So, as we know, when $n=1,$ we will get $n-1=0.$ Therefore, the derivative will become the following, \[\dfrac{dx}{dx}=1\times {{x}^{1-1}}=1\times {{x}^{0}}.\] We know that ${{x}^{0}}=1$ for any number $x.$ So, as a result of the above written identity, we will get the following derivative $\dfrac{dx}{dx}=1\times {{x}^{1-1}}=1\times {{x}^{0}}=1\times 1=1.$