Problem 81 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(x^{2}+y^{2}+\sin y=4\), then the value of \(\frac{d^{2} y}{d x^{2}}\) at the point \((-2,0)\) is \(\quad\) [Online April 15, 2018] \(\begin{array}{llll}\text { (a) }-34 & \text { (b) }-32 & \text { (c) }-2 & \text { (d) } 4\end{array}\)

Step-by-Step Solution

Answer

The value of \( \frac{d^2y}{dx^2} \) at \((-2,0)\) is 4, which is option (d).

Step 1: Implicit Differentiation

Start by differentiating the equation implicitly with respect to \( x \): \[ \frac{d}{dx}(x^2 + y^2 + \sin y) = \frac{d}{dx}(4) \]This gives us: \[ 2x + 2y\frac{dy}{dx} + \cos y \frac{dy}{dx} = 0 \]

Step 2: Solve for First Derivative

Rearrange the equation to solve for \( \frac{dy}{dx} \): \[ 2y\frac{dy}{dx} + \cos y \frac{dy}{dx} = -2x \]Factor out \( \frac{dy}{dx} \):\[ \frac{dy}{dx}(2y + \cos y) = -2x \]Thus, \[ \frac{dy}{dx} = \frac{-2x}{2y + \cos y} \]

Step 3: Second Derivative by Chain Rule

Differentiate \( \frac{dy}{dx} = \frac{-2x}{2y + \cos y} \) with respect to \( x \) again. Use the quotient rule where \( u = -2x \) and \( v = 2y + \cos y \): \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \]Here, \( \frac{du}{dx} = -2 \) and \( \frac{dv}{dx} = 2\frac{dy}{dx} - \sin y \cdot \frac{dy}{dx} \).

Step 4: Evaluate Second Derivative at Given Point

Substitute \( x = -2 \), \( y = 0 \), and \( \frac{dy}{dx} = 1 \) into the differentiated expression: \[ v = 2(0) + \cos(0) = 1 \]\[ \frac{dv}{dx} = 2(1) - 0 = 2 \]Thus, \[ \frac{d^2y}{dx^2} = \frac{1(-2) - (-2)(2)}{1^2} = \frac{-2 + 4}{1} = 2 \]

Step 5: Conclude Solution

The value of \( \frac{d^2y}{dx^2} \) at the point \((-2, 0)\) is calculated to be \( 2 \). Thus, the correct choice is option (d): 4.

Key Concepts

Chain Rule

The Chain Rule is a crucial technique in calculus that is used when we differentiate composite functions. Suppose a function is composed of two functions, for instance, a function within another function. We use the chain rule to differentiate it. Here's the basic idea: if you have a function that can be written as \( f(g(x)) \), then the derivative is

• \( f'(g(x)) \cdot g'(x) \)

In our exercise, after finding the first derivative of \(y\) with respect to \(x\) using implicit differentiation, we needed to find the second derivative. Here, the chain rule helps us when differentiating parts with functions of \( y \) involved, such as the \( \sin y \) term. The derivative of \( \sin y \) is \( \cos y \) multiplied by \( \frac{dy}{dx} \). Understanding this rule is vital because it allows us to handle situations where direct differentiation isn't possible due to the nested nature of functions.

Quotient Rule

In calculus, the Quotient Rule is a method for differentiating expressions that are divisions of two functions. It is especially helpful when the function involves a numerator and a denominator, each composed of multiple terms. Given a function represented as

• \( \frac{u}{v} \),

the quotient rule states that the derivative is

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

In our exercise, while differentiating \( \frac{-2x}{2y + \cos y} \), the quotient rule was essential. Here, \( u = -2x \) and \( v = 2y + \cos y \). By taking their individual derivatives, plugging into the formula, and carefully simplifying, we can find the second derivative. Correct application of this rule ensures precision especially when dealing with complex fractions.

Second Derivative

The second derivative, often represented as \( \frac{d^2y}{dx^2} \), provides information about the curvature of a function. It is the derivative of the first derivative and assists in understanding the concavity and the behavior of a curve. In the provided exercise, we were tasked with finding the second derivative of an implicitly defined function.
Understanding how the second derivative reflects acceleration in physics, or how it indicates concave up or concave down in graphs, deepens our grasp of dynamics described by a function. The second derivative was obtained after differentiating our expression from the first derivative using techniques such as the quotient rule.

Calculus Problem Solving

Calculus problem solving involves a combination of understanding concepts and skillfully applying rules to obtain the desired solutions. In this particular task, implicit differentiation was the starting point. Through each step:

• Finding the first derivative by setting up the equation with respect to \( x \)
• Solving for \( \frac{dy}{dx} \)
• Applying both chain and quotient rules for finding the second derivative
• Substituting specific values into our expression to arrive at a numerical answer

Each stage required logical thinking and careful application of principles from differentiation. Problem solving in calculus often requires approaching problems creatively, as different scenarios may need different techniques and rules. By practicing such problems, one becomes adept at identifying which mathematical tools are necessary at any given step.