Problem 68 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)=x+\tan x\) and \(f\) is the inverse of \(g\), then \(g^{\prime}(x)\) is equal to (A) \(\frac{1}{1+[g(x)-x]^{2}}\) (B) \(\frac{1}{2-[g(x)-x]^{2}}\) (C) \(\frac{1}{2+[g(x)-x]^{2}}\) (D) None of these

Step-by-Step Solution

Answer

(C) \( \frac{1}{2+[g(x)-x]^{2}} \)

Step 1: Understand the Relationship

Recognize that if \( f \) is the inverse of \( g \), then the composition of \( f \) and \( g \) satisfies \( f(g(x)) = x \) for all \( x \) in the domain.

Step 2: Apply the Chain Rule

Differentiate \( f(g(x)) = x \) with respect to \( x \). This gives \( f'(g(x)) \cdot g'(x) = 1 \). Solving for \( g'(x) \), we have \( g'(x) = \frac{1}{f'(g(x))} \).

Step 3: Find the Derivative of f(x)

Recall that \( f(x) = x + \tan x \). The derivative \( f'(x) = \frac{d}{dx}(x + \tan x) = 1 + \sec^2 x \).

Step 4: Substitute and Simplify

Substitute \( f'(g(x)) = 1 + \sec^2(g(x)) \) into \( g'(x) = \frac{1}{f'(g(x))} \). \( g'(x) = \frac{1}{1 + \sec^2(g(x))} \).

Step 5: Use Trigonometric Identity

Recall that \( \sec^2 x = 1 + \tan^2 x \), so \( 1 + \sec^2 x = 2 + \tan^2 x \). Substitute \( \tan g(x) = g(x) - x \) to get \( g'(x) = \frac{1}{2 + (g(x) - x)^2} \).

Step 6: Compare with the Options

Compare \( g'(x) = \frac{1}{2 + (g(x) - x)^2} \) with the given options. It matches option (C).

Key Concepts

Inverse Relationship

When two functions are inverses, there exists a special relationship between them. For functions \(f\) and \(g\) which are inverses, they undo each other's operations. This is expressed mathematically as \(f(g(x)) = x\) and \(g(f(x)) = x\).

• Imagine \(f\) transforming \(x\) into a new value, and then \(g\) taking that transformed value back to \(x\).
• This inverse relationship is fundamental in finding derivatives of inverse functions.

Furthermore, if \(f\) and \(g\) are inverse functions, their domains are the range of the other function. In the provided problem, since \(f(x) = x + \tan x\), \(g\) is its inverse, meaning it reverses the effect of \(f\).

Chain Rule

The chain rule is a key technique in calculus used when differentiating the composition of functions. Simply put, if you have a function \(F\) composed of an outer function \(f\) and an inner function \(g\), that is \(F(x) = f(g(x))\), then:
\[\frac{dF}{dx} = f'(g(x)) \cdot g'(x)\]This rule is essential for finding derivatives when dealing with inverse functions. In the problem, we used the chain rule to differentiate \(f(g(x)) = x\) with respect to \(x\).

• The derivative of \(x\) is \(1\).
• Using the chain rule, we got \(f'(g(x)) \cdot g'(x) = 1\).

This allows us to express \(g'(x)\) in terms of the derivative of \(f\), facilitating further simplifications.

Trigonometric Derivatives

Understanding trigonometric derivatives is a core part of calculus. Here, the derivative of a trigonometric function is pivotal in finding the inverse function's derivative. For instance, with \(f(x) = x + \tan x\):

• The derivative of \(x\) is \(1\).
• The derivative of \(\tan x\) is \(\sec^2 x\).

Thus, \(f'(x) = 1 + \sec^2 x\). This derivative informs us about the rate at which \(f(x)\) changes.
By recognizing the trigonometric identity \(\sec^2 x = 1 + \tan^2 x\), we can rewrite \(f'(g(x))\) as \(1 + \sec^2(g(x)) = 2 + \tan^2(g(x))\). This simplification is crucial when solving for \(g'(x)\).

Composition of Functions

Composition involves combining two functions such that the output of one function becomes the input to another. If \(f(x)\) and \(g(x)\) are functions, then the composition \(f(g(x))\) is a function itself.
In this exercise, \(f(x) = x + \tan x\) is a simple polynomial added to a trigonometric function. By applying \(f\) to \(g(x)\), we naturally resolve the inverse process because of the defined relationship \(f(g(x)) = x\).

• Composition highlights the chained processes between two functions.
• It's particularly important in understanding how transformations and their inverses interact.

To decode \(g'(x)\) from this composition, it's crucial to find expressions involving \(f'(g(x))\). This makes understanding compositions valuable for evaluating inverses and their derivatives.