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Question: The derivative of \(f\left( {{\tan }^{-1}}x \right),\) where \(f\left( x \right)=\tan x\) is \(\le...

The derivative of f(tan⁡−1x),f\left( {{\tan }^{-1}}x \right), where f(x)=tan⁡xf\left( x \right)=\tan x is
(a) 1\left( a \right)\,1
(b) 11+x2\left( b \right)\,\dfrac{1}{1+{{x}^{2}}}
(c) 2\left( c \right)\,2
(d) −11+x2\left( d \right)\,\dfrac{-1}{1+{{x}^{2}}}

Explanation

Solution

We will use some familiar trigonometric identity to solve this question. We will substitute for xx in the given definition of the function f.f. Then, we will use the trigonometric identity given by tan⁡(tan⁡−1x)=x.\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(tan⁡−1x).f\left( {{\tan }^{-1}}x \right).
We know that the function ff is defined as f(x)=tan⁡x.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 xx on the right-hand side.
So, f(x)f\left( x \right) will become f(tan⁡−1x)f\left( {{\tan }^{-1}}x \right) and so the right-hand side will become tan⁡(tan⁡−1x).\tan \left( {{\tan }^{-1}}x \right).
Thus, we will get the function as f(tan⁡−1x)=tan⁡(tan⁡−1x).f\left( {{\tan }^{-1}}x \right)=\tan \left( {{\tan }^{-1}}x \right).
We know the trigonometric identity given by tan⁡(tan⁡−1x)=x.\tan \left( {{\tan }^{-1}}x \right)=x.
So, when we substitute this in the definition of the function f,f, we will get f(tan⁡−1x)=x.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 ddxf(tan⁡−1x)=dxdx.\dfrac{d}{dx}f\left( {{\tan }^{-1}}x \right)=\dfrac{dx}{dx}.
We know that dxdx=1.\dfrac{dx}{dx}=1.
So, we will get the derivative of the function as ddxf(tan⁡−1x)=dxdx=1.\dfrac{d}{dx}f\left( {{\tan }^{-1}}x \right)=\dfrac{dx}{dx}=1.

Hence the derivative of the function f(tan⁡−1x)f\left( {{\tan }^{-1}}x \right) is equal to 11 where f(x)=tan⁡x.f\left( x \right)=\tan x.

Note:
We should always remember the trigonometric identities given by cos⁡(cos⁡−1x)=x\cos \left( {{\cos }^{-1}}x \right)=x and sin⁡(sin⁡−1x)=x.\sin \left( {{\sin }^{-1}}x \right)=x. We know the basic rule of differentiation which is given by dxndx=nxn−1.\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}. So, as we know, when n=1,n=1, we will get n−1=0.n-1=0. Therefore, the derivative will become the following, dxdx=1×x1−1=1×x0.\dfrac{dx}{dx}=1\times {{x}^{1-1}}=1\times {{x}^{0}}. We know that x0=1{{x}^{0}}=1 for any number x.x. So, as a result of the above written identity, we will get the following derivative dxdx=1×x1−1=1×x0=1×1=1.\dfrac{dx}{dx}=1\times {{x}^{1-1}}=1\times {{x}^{0}}=1\times 1=1.