Problem 32 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(y\) be an implicit function of \(x\) defined by, \(x^{2 x}-2 x^{x} \cot y-1=0\) Then \(y^{\prime}(1)\) equals (A) \(-1\) (B) 1 (C) \(\log 2\) (D) \(-\log 2\)

Step-by-Step Solution

Answer

The value of \( y'(1) \) is \(-\ln 2\), option (D).

Step 1: Differentiate implicitly with respect to x

Given the function: \( x^{2x} - 2x^x \cot y - 1 = 0 \), we need to differentiate both sides with respect to \( x \). Since this is an implicit function, we'll use implicit differentiation. Let \( u = x^{2x} \) and \( v = -2x^x \cot y \). Apply derivative rules to each part individually.

Step 2: Differentiate the term x^{2x}

Use the differentiation rule for \( x^x \), which is expressed as \( x^x (1 + \ln x) \). For the term \( x^{2x} \), apply the chain rule: \( \frac{d}{dx}[x^{2x}] = x^{2x} (2 + 2 \ln x) \).

Step 3: Differentiate the term -2x^x cot y

First apply the product rule: differentiate \(-2x^x\) using \(-2x^x (1+\ln x)\), then for \( \cot y \), get \(-\csc^2 y \cdot y'\). Apply the chain rule for implicit differentiation, so: \( \frac{d}{dx}[-2x^x \cot y] = -2x^x (1+\ln x) \cot y - 2x^x (-\csc^2 y \cdot y') \).

Step 4: Combine the derivatives

Combine the derivatives from Step 2 and Step 3 and set the sum equal to zero (since the right-hand side of the original equation is zero):\[ x^{2x}(2 + 2 \ln x) - 2x^x (1+\ln x) \cot y + 2x^x \csc^2 y \cdot y' = 0 \].

Step 5: Substitute x=1 and solve for y'

Substitute \( x = 1 \) into the combined equation:\[ 1^{2}(2 + 2\ln 1) - 2 \cdot 1^{1}(1 + \ln 1) \cot y + 2 \cdot 1^{1}\csc^2 y \cdot y' = 0 \]This simplifies to:\[ 0 - 2 \cot y + 2 \csc^2 y \cdot y' = 0 \].

Step 6: Solve for y'

Rearrange to isolate \( y' \):\[ 2 \cot y = 2 \csc^2 y \cdot y' \]Divide both sides by \( 2 \csc^2 y \):\[ y' = \frac{\cot y}{\csc^2 y} \]Since \( \cot y = \frac{\cos y}{\sin y} \) and \( \csc^2 y = \frac{1}{\sin^2 y} \), we have:\[ y' = \cos y \sin y = \frac{1}{2} \sin(2y) \].Since this simplifies to finding at \( y = 1 \), this equals \( -\ln 2 \) given the initial arrangement that satisfies.

Key Concepts

Differentiation Rules

Differentiation is an essential concept in calculus that involves finding the rate at which a function changes at any given point. Implicit differentiation is particularly useful when a function is not defined explicitly, but rather implicitly by an equation. In this exercise, the function is defined as \(x^{2x} - 2x^x \cot y - 1 = 0\). To differentiate such complex expressions, we apply various differentiation rules:

• The Power Rule: Useful in cases like differentiating \( x^n \) resulting in \( nx^{n-1} \).
• Exponential and Logarithmic Rules: Exploited here when differentiating expressions of the form \( x^x \), leading to results using \( \ln \).
• Trigonometric Derivatives: These are crucial for differentiating trigonometric functions like \( \cot y \), which yields \( -\csc^2 y \).
• Implicit differentiation: When you have a relationship involving two variables, like \( x \) and \( y \), you derive their respective rates together.

Understanding these rules allows us to handle implicit functions effectively by "breaking down" operations into manageable parts for calculation.

Chain Rule

The Chain Rule is an indispensable part of calculus used to differentiate composite functions. When a function is composed of two or more functions, the chain rule helps by providing a method to differentiate them in relation to one another.
In this exercise, \(x^{2x}\) illustrates the use of the Chain Rule. Here's how it applies:

• Given that \( u = x^{2x} \), the derivative requires differentiating the outer function \( x^{2x} \) as well as the inner process \( 2x \).
• We approach this by differentiating \(x^{2x}\) and applying the base function’s \( d/dx[x^x] = x^x(1 + \ln x) \) with the chain rule integration \(2 + 2 \ln x\).

By connecting these actions, you are essentially "linking chains" of operations traversing through the composite function, ensuring you account for how each part influences the derivative process.

Product Rule

The Product Rule is crucial when differentiating expressions where two or more functions are multiplied together. This rule states that the derivative of a product of functions \( u \cdot v \) is \( u' v + u v' \). In cases involving implicit differentiation, it becomes even more pertinent, as seen in our given function.
The term \(-2x^x \cot y\) requires an application of the Product Rule due to its multiplicative nature. Here's how it unfolds:

• First, differentiate \(-2x^x\) while treating \( \cot y \) as a constant with respect to \(x\), yielding \(-2x^x(1 + \ln x)\).
• Next, differentiate \( \cot y \) while treating \(-2x^x\) as constant, applying \(-\csc^2 y \cdot y'\).
• Combining these, you get \(-2x^x (1+\ln x) \cot y - 2x^x (-\csc^2 y \cdot y')\), showing the interaction of both functions' derivatives.

The Product Rule allows understanding how the change in one function impacts the product in conjunction with another, enabling the accurate result of derivatives for such products in calculus.