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

Question

If \(\mathrm{e}^{y}+x y=e\), the ordered pair \(\left(\frac{d y}{d x}, \frac{d^{2} y}{d x^{2}}\right)\) at \(x=0\) is equal to: (a) \(\left(\frac{1}{e},-\frac{1}{e^{2}}\right)\) (b) \(\left(-\frac{1}{e}, \frac{1}{e^{2}}\right)\) (c) \(\left(\frac{1}{e}, \frac{1}{e^{2}}\right)\) (d) \(\left(-\frac{1}{e},-\frac{1}{e^{2}}\right)\)

Step-by-Step Solution

Answer

(b) \(\left(-\frac{1}{e}, \frac{1}{e^2}\right)\)

Step 1: Differentiate the equation with respect to x

Given the equation \( e^y + x y = e \), we need to differentiate both sides with respect to \( x \). Using the chain rule, the derivative of \( e^y \) is \( e^y \frac{dy}{dx} \) and the derivative of \( xy \) is \( y + x \frac{dy}{dx} \). Thus, differentiating both sides gives: \[ e^y \frac{dy}{dx} + y + x \frac{dy}{dx} = 0. \] Rearrange to solve for \( \frac{dy}{dx} \): \[ \left(e^y + x \right) \frac{dy}{dx} = -y. \] Then, \( \frac{dy}{dx} = \frac{-y}{e^y + x}. \)

Step 2: Evaluate at x = 0

Now, plug \( x = 0 \) into the differentiated equation. The original equation \( e^y + x y = e \) at \( x = 0 \) reduces to \( e^y = e \), which implies \( y = 1 \). Substitute \( y = 1 \) and \( x = 0 \) into the derivative \( \frac{dy}{dx} = \frac{-y}{e^y + x} \), giving \( \frac{dy}{dx} = \frac{-1}{e} \).

Step 3: Differentiate the first derivative with respect to x for second derivative

Differentiate \( \frac{dy}{dx} = \frac{-y}{e^y + x} \) with respect to \( x \) to find \( \frac{d^2y}{dx^2} \). This requires the quotient rule, where \( u = -y \) and \( v = e^y + x \). The quotient rule states: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}. \]

Step 4: Apply the quotient rule

For \( u = -y \), we have \( \frac{du}{dx} = -\frac{dy}{dx} = \frac{1}{e} \) (from Step 2). For \( v = e^y + x \), we differentiate to get \( \frac{dv}{dx} = e^y \frac{dy}{dx} + 1 \). Substitute \( y = 1 \), \( x = 0 \), and \( \frac{dy}{dx} = \frac{-1}{e} \) into these expressions: \( \frac{dv}{dx} = e \cdot \left(\frac{-1}{e}\right) + 1 = -1 + 1 = 0 \). Substitute back into the quotient rule formula: \[ \frac{d^2 y}{dx^2} = \frac{(e + 0) \left(\frac{1}{e}\right) - (-1)(0)}{(e + 0)^2} = \frac{e \cdot \frac{1}{e}}{e^2} = \frac{1}{e^2}. \]

Key Concepts

Chain Rule

In calculus, the Chain Rule is a fundamental technique used to differentiate composite functions. When you have a function nested inside another function, the Chain Rule helps you differentiate it efficiently. For example, if you have a function of the form \( f(g(x)) \), the derivative is found using the Chain Rule:

• First, differentiate the outer function \( f \) as if the inner function \( g(x) \) is just a variable.
• Then, multiply the result by the derivative of the inner function \( g(x) \).

This rule was applied in the solution when differentiating \( e^y \). Here, \( y \) is treated as a function of \( x \) (i.e., \( y = g(x) \)). Thus, the derivative of \( e^y \) involved applying the Chain Rule, resulting in \( e^y \cdot \frac{dy}{dx} \). Breaking functions into parts like this helps simplify the process and leads to a clearer understanding and solution.

Quotient Rule

The Quotient Rule is a method used in differentiation when dealing with functions that are expressed as fractions. When you have a function \( f(x) = \frac{u(x)}{v(x)} \), where both \( u \) and \( v \) are differentiable functions, the Quotient Rule states:

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

In the step-by-step solution, this rule was crucial when finding the second derivative \( \frac{d^2y}{dx^2} \). Here, the function \( \frac{-y}{e^y + x} \) was differentiated by identifying \( u = -y \) and \( v = e^y + x \).

• First, calculate \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).
• Then apply the quotient rule formula.

Understanding how to navigate and apply the Quotient Rule allows solving even complex situations like rational expressions within derivatives.

Partial Derivatives

Partial derivatives are used in multivariable calculus, where functions depend on more than one variable. These derivatives focus on how a function changes as one of the variables changes, holding the others constant. In other words, a partial derivative measures the rate of change of a function with respect to one variable while keeping other variables fixed. While this specific exercise didn't dive into partial derivatives directly, understanding them helps grasp the broader scope of differentiation techniques. When working on problems involving multiple variables, comprehension of partial derivatives becomes pivotal in fields like optimization and physics.

• When expressing, use the notation \( \frac{\partial f}{\partial x} \) to indicate the partial derivative of a function \( f \) with respect to the variable \( x \).
• This approach ensures clarity when there are multiple variables involved, each examined in isolation while others stay constant.

Gaining skills in partial derivatives will prepare you for more advanced calculus topics and enhance your analytical abilities.