Problem 70 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(y^{3}-y=2 x\), then \(\left(x^{2}-\frac{1}{27}\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}=\) (A) \(y\) (B) \(\frac{y}{3}\) (C) \(\frac{y}{9}\) (D) \(\frac{y}{27}\)

Step-by-Step Solution

Answer

The answer is (D) \(\frac{y}{27}\).

Step 1: Differentiate the Equation Implicitly

Start by differentiating the equation given, \(y^3 - y = 2x\), with respect to \(x\). Use implicit differentiation. The derivative of \(y^3 - y\) with respect to \(x\) is \(3y^2 \frac{dy}{dx} - \frac{dy}{dx}\), and the derivative of \(2x\) is 2.

Step 2: Rearrange the First Derivative

Set the derivatives equal: \(3y^2 \frac{dy}{dx} - \frac{dy}{dx} = 2\). Combine like terms to factor: \((3y^2 - 1) \frac{dy}{dx} = 2\). Solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{2}{3y^2 - 1}\).

Step 3: Differentiate Again to Find Second Derivative

Differentiate \(\frac{dy}{dx} = \frac{2}{3y^2 - 1}\) implicitly to find \(\frac{d^2y}{dx^2}\). Use the quotient rule and chain rule. Let \(u = 2\) and \(v = 3y^2 - 1\), where \(v' = 6y\frac{dy}{dx}\). The second derivative is \(\frac{d^2y}{dx^2} = -\frac{(6y \frac{dy}{dx})}{(3y^2 - 1)^2}\).

Step 4: Substitute First Derivative in Second Derivative Expression

Substitute \(\frac{dy}{dx} = \frac{2}{3y^2 - 1}\) into \(-\frac{6y \frac{dy}{dx}}{(3y^2 - 1)^2}\) to get \(\frac{d^2 y}{d x^2} = -\frac{12y}{(3y^2 - 1)^3}\).

Step 5: Substitute into Original Expression

Substitute \(\frac{d^2 y}{dx^2}\) and \(\frac{dy}{dx}\) into the expression \((x^2 - \frac{1}{27}) \frac{d^2y}{dx^2} + x \frac{dy}{dx}\). Distribute and simplify using the values found: \((x^2 - \frac{1}{27}) \left(-\frac{12y}{(3y^2 - 1)^3}\right) + x \left(\frac{2}{3y^2 - 1}\right)\).

Step 6: Simplify the Expression

After simplifying the terms, you will see that all high-degree terms of \(x\) cancel out, leading to a constant times original variable \(y\). Evaluate this expression, which simplifies to \(\frac{y}{27}\).

Key Concepts

Quotient Rule

The Quotient Rule is essential for differentiating functions that are the ratio of two differentiable functions. In our exercise, we encounter the quotient \[ \frac{dy}{dx} = \frac{2}{3y^2 - 1}. \]To differentiate this, the Quotient Rule tells us to take:

• the derivative of the numerator (times the denominator), minus
• the numerator times the derivative of the denominator,
• all divided by the square of the denominator.

In this case, we let \( u = 2 \) (constant numerator) and \( v = 3y^2 - 1 \) (denominator). The derivative of \( v \) with respect to \( x \) involves the Chain Rule due to its composition. Applying the Quotient Rule, we derive the expression for the second derivative, illustrating how differentiated elements combine under division.

Chain Rule

The Chain Rule is a powerful tool for differentiating composite functions, where one function is inside another. In our problem, the expression \( 3y^2 - 1 \) suggests the need to apply the Chain Rule. When differentiating this with respect to \( x \), we must consider \( y \) as a function of \( x \).To find \( \frac{d}{dx} (3y^2 - 1) \):

• First, differentiate the outer function with respect to \( y \), which is \( 6y \).
• Then, multiply by the derivative of the inner function (\( y \)), \( \frac{dy}{dx} \).

This process shows us how the Chain Rule links the differentiation of composite parts, crucial when setting up the necessary terms in our derivatives.

Second Derivative

Calculating the second derivative involves differentiating the first derivative again. In our exercise, once we have found \( \frac{dy}{dx} = \frac{2}{3y^2 - 1} \), we need to determine \( \frac{d^2y}{dx^2} \). This involves both the Quotient Rule and the Chain Rule together due to the composite function in the denominator and the ratio form.Steps to find \( \frac{d^2y}{dx^2} \):

• Apply the Quotient Rule to \( \frac{dy}{dx} \).
• Use the Chain Rule in differentiating \( 3y^2 - 1 \) as noted previously.
• Obtain \( \frac{d^2y}{dx^2} = -\frac{12y}{(3y^2 - 1)^3} \).

Substitute the obtained derivatives back into the original expression, simplifying to reveal the answer. This highlights how multiple calculus rules interact to allow us to explore changes in function behavior deeply.