Question

If f(x) = x + tan x and f is the inverse of g, then g’(x) is equal to
\[\left( \text{a} \right)\text{ }\dfrac{1}{1+{{\left[ g\left( x \right)-x \right]}^{2}}}\]
\[\left( \text{b} \right)\text{ }\dfrac{1}{2-{{\left[ g\left( x \right)+x \right]}^{2}}}\]
\[\left( \text{c} \right)\text{ }\dfrac{1}{2+{{\left[ x-g\left( x \right) \right]}^{2}}}\]
(d) None of these

Answer 1

Hint: To solve the given question, we will first find out what an inverse function is. Then we will use the concept that if f is inverse of g, then we can write, \[g\left( x \right)={{f}^{-1}}\left( x \right)\] which will give us \[f\left( g\left( x \right) \right)=x.\] We will find out the value of \[f\left( g\left( x \right) \right)\] by putting g (x) in place of x in the equation \[f\left( x \right)=x+\tan x.\] After doing this, we will differentiate \[f\left( g\left( x \right) \right)=x\] on both the sides. While differentiating this, we will use the chain rule on the left-hand side and then we will find the value of g’(x) in terms of g(x).

Complete step-by-step answer:
Before we solve the question given, we must know what an inverse function is. An inverse function is a function that undoes the action of another function. A function U is the inverse of a function V if whenever y = U(x) then, x = V(y).
Now, in the question, we are given that f is inverse of g. Thus, we can write this as,
\[g\left( x \right)={{f}^{-1}}\left( x \right)\]
\[\Rightarrow f\left( g\left( x \right) \right)=x.....\left( i \right)\]
Now, we will find the value of \[f\left( g\left( x \right) \right).\] For this, we will put g(x) in place of x in the equation \[f\left( x \right)=x+\tan x.\] Thus, we will get the following equation
\[\Rightarrow f\left( g\left( x \right) \right)=g\left( x \right)+\tan \left( g\left( x \right) \right)....\left( ii \right)\]
Now, we will put the value of \[f\left( g\left( x \right) \right)\] from (ii) to (i). Thus, we will get,
\[g\left( x \right)+\tan \left( g\left( x \right) \right)=x.....\left( iii \right)\]
Now, we will differentiate both sides of the equation (iii) with respect to x. Thus, we will get,
\[\dfrac{d}{dx}\left[ g\left( x \right)+\tan \left( g\left( x \right) \right) \right]=\dfrac{d}{dx}\left( x \right)\]
\[\Rightarrow \dfrac{d}{dx}\left( g\left( x \right) \right)+\dfrac{d}{dx}\left[ \tan \left( g\left( x \right) \right) \right]=\dfrac{d}{dx}\left( x \right)\]
Now, the differentiation of g(x) and x with respect to x will be g’(x) and 1 respectively. Thus, we will get,
\[\Rightarrow {{g}^{'}}\left( x \right)+\dfrac{d}{dx}\left[ \tan \left( g\left( x \right) \right) \right]=1\]
For the differentiation of tan (g(x)) with respect to x, we will use the chain rule. The chain rule says that differentiation A(B(x)) is given by
\[\dfrac{d}{dx}\left[ A\left( B\left( x \right) \right) \right]={{A}^{'}}\left( B\left( x \right) \right)\times {{B}^{'}}\left( x \right)\]
Using this, we will get,
\[{{g}^{'}}\left( x \right)+{{\sec }^{2}}g\left( x \right).{{g}^{'}}\left( x \right)=1.....\left( iv \right)\]
In (iv), we have used \[\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x.\] On taking g’(x) common, we will get,
\[{{g}^{'}}\left( x \right)\left[ 1+{{\sec }^{2}}\left( g\left( x \right) \right) \right]=1\]
\[\Rightarrow {{g}^{'}}\left( x \right)=\dfrac{1}{1+{{\sec }^{2}}g\left( x \right)}\]
Now, we know the identity that \[{{\sec }^{2}}\theta ={{\tan }^{2}}\theta +1.\] Thus, we will get,
\[\Rightarrow {{g}^{'}}\left( x \right)=\dfrac{1}{1+{{\tan }^{2}}g\left( x \right)+1}\]
\[\Rightarrow {{g}^{'}}\left( x \right)=\dfrac{1}{2+{{\tan }^{2}}g\left( x \right)}......\left( v \right)\]
From (iii), we have,
\[\Rightarrow g\left( x \right)+\tan \left( g\left( x \right) \right)=x\]
\[\Rightarrow \tan \left( g\left( x \right) \right)=x-g\left( x \right)\]
On squaring both the sides, we will get,
\[\Rightarrow {{\tan }^{2}}g\left( x \right)={{\left( x-g\left( x \right) \right)}^{2}}.....\left( vi \right)\]
On putting the value of \[{{\tan }^{2}}g\left( x \right)\] from (vi) to (v), we will get,
\[{{g}^{'}}\left( x \right)=\dfrac{1}{2+{{\left( x-g\left( x \right) \right)}^{2}}}\]
Hence, option (c) is the right answer

Note: We can find the inverse of this function because the given function f(x) is a bijective function. This means f(x) is one – one and onto function. This property is necessary for the function to have an inverse. If f(x) would have been non – bijective, then option (d) would have been correct.