Problem 75 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

For \(x>1\), if \((2 x)^{2 y}=4 e^{2 x-2 y}\), then \(\left(1+\log _{e} 2 x\right)^{2} \frac{d y}{d x}\) is equal to (a) \(\frac{x \log _{e} 2 x-\log _{e} 2}{x}\) (b) \(\log _{e} 2 x\) (c) \(\frac{x \log _{e} 2 x+\log _{e} 2}{x}\) (d) \(x \log _{e} 2 x\)

Step-by-Step Solution

Answer

The answer is (a) \(\frac{x \log_e 2x - \log_e 2}{x}\).

Step 1: Simplify the given equation

Rewrite the given equation \((2x)^{2y} = 4e^{2x - 2y}\). Since \(4\) can be expressed as \((2)^2\), rewrite the equation as: \((2x)^{2y} = (2)^2 e^{2x - 2y}\).

Step 2: Apply logarithm to both sides

Take the natural logarithm (\(\log_e\)) of both sides to simplify further. By applying logarithmic properties: \[\ln((2x)^{2y}) = \ln((2)^2e^{2x - 2y})\]This simplifies to: \[2y \ln(2x) = 2\ln(2) + (2x - 2y)\]

Step 3: Rearrange and differentiate

Re-arrange the equation to solve for \(y\):\[2y \ln(2x) + 2y = 2\ln(2) + 2x\]Now, differentiate both sides with respect to \(x\) using implicit differentiation: \[\frac{d}{dx}(2y \ln(2x) + 2y) = \frac{d}{dx}(2\ln(2) + 2x)\]

Step 4: Differentiate both sides

Apply the product rule to the left-hand side and differentiate:\[2 \left( \ln(2x) \cdot \frac{dy}{dx} + y \cdot \frac{1}{x} \right) + 2 \cdot \frac{dy}{dx} = 2\]Which simplifies to:\[2 \ln(2x) \frac{dy}{dx} + \frac{2y}{x} + 2 \frac{dy}{dx} = 2\]

Step 5: Solve for \(\frac{dy}{dx}\)

Factor out \(\frac{dy}{dx}\):\[\frac{dy}{dx}(2 \ln(2x) + 2) = 2 - \frac{2y}{x}\]Solving for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{2 - \frac{2y}{x}}{2 (\ln(2x) + 1)}\]

Step 6: Solve for \(y\) with initial condition and express final equation

Use the rearranged equation from Step 3:\[y (\ln(2x) + 1) = x + \ln(2)\]Express \(y\) as:\[y = \frac{x + \ln(2)}{\ln(2x) + 1}\]

Step 7: Substitute \(y\) back into \(\frac{dy}{dx}\)

Substitute the expression for \(y\) into \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{2 - \frac{2(x + \ln(2))/(\ln(2x) + 1)}{x}}{2(\ln(2x) + 1)}\]This simplifies to:\[\frac{dy}{dx} = \frac{x \ln(2x) - \ln(2)}{x(\ln(2x) + 1)}\]

Step 8: Find \((1 + \ln(2x))^{2} \frac{dy}{dx}\) and choose the correct answer

Multiply \(\frac{dy}{dx}\) by \((1 + \ln(2x))^2\): \[(1 + \ln(2x))^2 \frac{dy}{dx} = (1 + \ln(2x)) \left( \frac{x \ln(2x) - \ln(2)}{x} \right) = \frac{x \ln(2x) - \ln(2)}{x}\]This matches option (a).

Key Concepts

Implicit Differentiation

Understanding implicit differentiation is crucial to solving equations where the dependent variable, typically denoted as \(y\), is not isolated on one side of the equation. Instead, \(y\) appears intertwined with \(x\) in a more complex expression. In such cases, traditional differentiation approaches fall short. Implicit differentiation steps in as a method to differentiate both sides of an equation with respect to \(x\). This method, in essence, treats \(y\) as a function of \(x\) even when it's not explicitly given.

To apply implicit differentiation, follow these steps:

• Differentiate every term of the equation with respect to \(x\).
• Remember to apply the chain rule when differentiating terms involving \(y\), since \(y\) is a function of \(x\) (i.e., each time you differentiate \(y\), multiply by \(\frac{dy}{dx}\)).
• Solve for \(\frac{dy}{dx}\) after differentiating.

In our exercise, implicit differentiation allowed us to differentiate both sides after rearranging terms. This process enabled us to successfully derive the expression for \(\frac{dy}{dx}\).

Logarithmic Properties

When dealing with complicated equations, like the one in the exercise, logarithmic properties can simplify our work remarkably. Logarithms have unique properties that transform multiplication into addition, powers into products, and more.

Key logarithmic properties to remember include:

• \(\log_b(mn) = \log_b(m) + \log_b(n)\)
• \(\log_b(m^n) = n \cdot \log_b(m)\)
• \(\log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n)\)

For our problem, applying \(\ln\) (the natural logarithm, where base \(e\) is used) to both sides simplified the equation. Specifically, it streamlined our differentiation process, making it feasible to isolate \(y\) in terms of \(x\).

This facilitation is central to solving such problems efficiently in calculus, emphasizing the importance of mastering logarithmic properties.

Calculus

Calculus forms the backbone of tackling this exercise. It involves understanding derivatives, particularly within the context of implicit differentiation, and applying chain and product rules accurately. The solution leverages calculus principles by:

• Using the derivative rules to find \(\frac{dy}{dx}\).
• The product rule is specifically vital when dealing with expressions involving products of functions, as seen in the differentiation part of our solution.
• The chain rule helps to differentiate composite functions, particularly when involving \(y\), an implied function of \(x\).

In the realm of calculus problems, grasping these rules is essential to transgressing more complex differentiations and integrations. This proficiency aids in recognizing patterns and correctly applying relevant formulas and properties, as displayed in the problem-solving steps of our given exercise.