Question

If g is the inverse of the function f(x) and \( g(x) = x + \tan x \) then, \( f'(x) = \)

• A. \( 1+\sec^2x \)
• B. \( \frac{1}{1+\sec^2f(x)} \)
• C. \( \frac{1}{1+\sec^2g(x)} \)
• D. \( 1+\sec^2f(x) \)

Hint:

If \(g\) is the inverse of \(f\), then \(f(g(x))=x\) and \(g(f(x))=x\). Differentiating \(g(f(x))=x\) w.r.t \(x\): \(g'(f(x)) \cdot f'(x) = 1 \implies f'(x) = \frac{1}{g'(f(x))}\). Alternatively, if \(y=f(x)\), then \(x=g(y)\). Then \( \frac{dy}{dx} = \frac{1}{dx/dy} \). Calculate \( \frac{dx}{dy} \) from \(x=g(y)\) and then substitute \(y=f(x)\). Given \(g(x) = x+\tan x\). If we say \(y=f(x)\), then \(x=g(y)=y+\tan y\). \( \frac{dx}{dy} = g'(y) = 1+\sec^2y \). So \( f'(x) = \frac{dy}{dx} = \frac{1}{1+\sec^2y} = \frac{1}{1+\sec^2(f(x))} \).

Answer 1

The Correct Option is B

Given \(g(x)\) is the inverse of \(f(x)\), so \(g(f(x)) = x\) and \(f(g(x)) = x\).
We are given \(g(x) = x + \tan x\).
We need to find \(f'(x)\).
We know the formula for the derivative of an inverse function: \( f'(x) = \frac{1}{g'(f(x))} \).
First, find \(g'(x)\).
Given \(g(x) = x + \tan x\).
\[ g'(x) = \frac{d}{dx}(x + \tan x) = 1 + \sec^2 x \] Now, substitute this into the formula for \(f'(x)\): \[ f'(x) = \frac{1}{g'(f(x))} \] Replace \(x\) with \(f(x)\) in the expression for \(g'(x)\): \[ g'(f(x)) = 1 + \sec^2(f(x)) \] Therefore, \[ f'(x) = \frac{1}{1 + \sec^2(f(x))} \] This matches option (2).
Let's verify the variable.
The question asks for \(f'(x)\).
Option (3) is \( \frac{1}{1+\sec^2g(x)} \).
If we differentiate \( f(g(x)) = x \) with respect to \(x\): \( f'(g(x)) \cdot g'(x) = 1 \).
So \( f'(g(x)) = \frac{1}{g'(x)} \).
If \(y = g(x)\), then \( f'(y) = \frac{1}{g'(x)} = \frac{1}{g'(g^{-1}(y))} = \frac{1}{g'(f(y))} \).
Replacing \(y\) with \(x\), we get \( f'(x) = \frac{1}{g'(f(x))} \).
This is consistent.
The variable in the options is \(x\).
Option (2) is \( \frac{1}{1+\sec^2f(x)} \).
Option (3) has \(g(x)\) in the denominator argument.
If the question was asking for \( (g^{-1})'(x) \) where \(g^{-1} = f \), this is correct.
Let \( y = f(x) \).
Then \( x = g(y) = y + \tan y \).
We want \( \frac{dy}{dx} \).
We have \( \frac{dx}{dy} \).
\( \frac{dx}{dy} = \frac{d}{dy}(y+\tan y) = 1 + \sec^2 y \).
So, \( f'(x) = \frac{dy}{dx} = \frac{1}{dx/dy} = \frac{1}{1+\sec^2 y} \).
Since \( y=f(x) \), we have \[ f'(x) = \frac{1}{1+\sec^2(f(x))} \] This confirms option (2).