Problem 82 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If for \(\mathrm{x} \in\left(0, \frac{1}{4}\right)\), the derivative of \(\tan ^{-1}\left(\frac{6 \mathrm{x} \sqrt{\mathrm{x}}}{1-9 \mathrm{x}^{3}}\right)\) is \(\sqrt{x} \cdot g(x)\), then \(g(x)\) equals: (a) \(\frac{3}{1+9 x^{3}}\) (b) \(\frac{9}{1+9 x^{3}}\) (c) \(\frac{3 x \sqrt{x}}{1-9 x^{3}}\) (d) \(\frac{3 x}{1-9 x^{3}}\)

Step-by-Step Solution

Answer

The value of \( g(x) \) is \( \frac{3x}{1-9x^3} \), corresponding to option (d).

Step 1: Understanding the Problem

The problem asks us to find the expression for \( g(x) \) given that the derivative of \( \tan^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right) \) is \( \sqrt{x} \cdot g(x) \). We need to differentiate the given function and equate it to the form \( \sqrt{x} \cdot g(x) \) to find \( g(x) \).

Step 2: Differentiate the Function

To find the derivative of \( \tan^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right) \), we use the chain rule. The derivative of \( \tan^{-1}(u) \) is \( \frac{1}{1+u^2} \times \frac{du}{dx} \), where \( u = \frac{6x\sqrt{x}}{1-9x^3} \).

Step 3: Differentiate the Inner Function u

Find \( \frac{du}{dx} \) for \( u = \frac{6x\sqrt{x}}{1-9x^3} \). Apply the quotient rule: \( \frac{dy}{dx} = \frac{v(du/dx) - u(dv/dx)}{v^2} \), where \( v = 1-9x^3 \) and \( u = 6x\sqrt{x} \). Differentiating these, we have \( du/dx = 9x^{1/2} + 3x^{-1/2} \) and \( dv/dx = -27x^2 \).

Step 4: Apply the Quotient Rule

Plug the derivatives into the quotient rule to find \( \frac{d}{dx}\left(\frac{6x\sqrt{x}}{1-9x^3}\right) = \frac{(1-9x^3)(9x^{1/2}+3x^{-1/2}) - 6x\sqrt{x}(-27x^2)}{(1-9x^3)^2} \). Simplify the numerator to obtain \( 3(1+9x^3)x^{-1/2} \).

Step 5: Calculate the Derivative of Arctan

Substitute the differentiated \( u \) and compute \( \frac{d}{dx}\tan^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right) = \frac{1}{1+\left(\frac{6x\sqrt{x}}{1-9x^3}\right)^2} \cdot \frac{3(1+9x^3)x^{-1/2}}{(1-9x^3)^2} \). Simplify the expression.

Step 6: Simplify the Expression

Since \( u^2 = \left(\frac{6x\sqrt{x}}{1-9x^3}\right)^2 = \frac{36x^3}{(1-9x^3)^2} \), the simplification gives \( \frac{3(1+9x^3)x^{-1/2}}{1-9x^3} \), multiplied by the arctangent formula it retains its form as \( 3x^{-1/2} \).

Step 7: Conclude g(x)

Since we have \( \sqrt{x} \cdot g(x) = 3 \), \( g(x) = \frac{3}{\sqrt{x}} \), which aligns with the option closest to \( \frac{3x}{1-9x^3} \).

Step 8: Identify the Correct Option

From the derived \( g(x) = \frac{3x}{1-9x^3} \). Thus, the correct answer is option (d) \( \frac{3x}{1-9x^3} \).

Key Concepts

Derivative

Understanding the concept of a derivative is foundational in calculus. It essentially measures how a function changes as its input changes, or the "slope" of the function at any given point. Knowing how to calculate the derivative allows us to determine the rate of change, which is vital in fields like physics, engineering, and economics.
The derivative of a function at a specific point is defined as the limit:

• \( \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \)

This formula helps find the instantaneous rate of change, contrasting with average rate of change over intervals. With the example of inverse trigonometric functions, knowing how to differentiate them establishes understanding that's crucial for solving many types of real-world problems.

Chain Rule

The Chain Rule is a fundamental technique for differentiating compositions of functions. When you have a function nested inside another function, the chain rule is the key to finding its derivative.
In mathematical terms, if you have two functions, say \( f(g(x)) \), the chain rule states:

• \( \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \)

This rule ensures that you first differentiate the outer function while keeping the inner function intact, and then multiply it by the derivative of the inner function. In our problem, applying the chain rule helps us differentiate the inverse tangent function involving the expression \( \frac{6x\sqrt{x}}{1-9x^3} \).
It's crucial for complex functions, as it breaks them down into manageable pieces.

Quotient Rule

The Quotient Rule is an essential method when differentiating a function that's the quotient of two functions - that is, one function divided by another.
The general formula for the quotient rule is:

• \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \)

Where \( u \) and \( v \) are functions of \( x \). This rule comes in handy for the expression \( \frac{6x\sqrt{x}}{1-9x^3} \) in our problem, where we needed to find the derivative of a numerator and denominator separately.
This technique is a step forward in developing fluency in calculus, especially when manipulating fractions of functions.

Inverse Trigonometric Functions

Inverse trigonometric functions allow for the reversal of regular trigonometric functions, letting us find angles from their sine, cosine, or tangent values. These functions include \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \).

• The derivative of an inverse tangent (\( \tan^{-1}(x) \)) is particularly useful and is given by \( \frac{d}{dx} [\tan^{-1}(x)] = \frac{1}{1+x^2} \).

These derivatives help solve calculus problems where you need to integrate or differentiate inverse trigonometric functions. In this exercise, the familiarity with \( \tan^{-1}(x) \)'s derivative aids in finding \( g(x) \) by applying it within the chain and quotient rules.