Problem 106 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(y=\left(x+\sqrt{1+x^{2}}\right)^{\mathrm{n}}\), then \(\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}\) is : [2002] (A) \(n^{2} y\) (B) \(-n^{2} y\) (C) \(-y\) (D) \(2 x^{2} y\)

Step-by-Step Solution

Answer

The correct answer is (A) \( n^2 y \).

Step 1: Express Derivatives Implicitly

Recognize that the expression for \( y \) can be differentiated implicitly with respect to \( x \). The idea is to find \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \).

Step 2: Differentiate with Exponent Rule

Using the fact that \( y = (x + \sqrt{1+x^2})^n \), apply the chain rule for differentiation: \( \frac{dy}{dx} = n(x + \sqrt{1+x^2})^{n-1} \cdot \frac{d}{dx}(x + \sqrt{1+x^2}) \).

Step 3: Derivative of Inner Function

Calculate \( \frac{d}{dx}(x + \sqrt{1+x^2}) \). This gives: \( 1 + \frac{x}{\sqrt{1+x^2}} \).

Step 4: Combine First Derivative

Substitute the derivative of the inner function back into the expression for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = n(x + \sqrt{1+x^2})^{n-1} \left(1 + \frac{x}{\sqrt{1+x^2}}\right) \].

Step 5: Differentiate Again for Second Derivative

Compute \( \frac{d^2y}{dx^2} \) by differentiating \( \frac{dy}{dx} \) again. This will involve the product rule and differentiating the exponents and terms involved.

Step 6: Use Product Rule on Second Derivative

Differentiate \( \frac{dy}{dx} \) considering it as a product of two functions: Use \( (f \cdot g)' = f'g + fg' \) where \( f = n(x + \sqrt{1+x^2})^{n-1} \) and \( g = 1 + \frac{x}{\sqrt{1+x^2}} \).

Step 7: Assemble Second Derivative Expression

From differentiating \( \frac{dy}{dx} \) again, bundle terms appropriately and express \( \frac{d^2y}{dx^2} \) as a simplified function.

Step 8: Substitute into Given Equation

Substitute \( \frac{d^2y}{dx^2} \) and \( \frac{dy}{dx} \) into the equation \( (1 + x^2) \frac{d^2y}{dx^2} + x \frac{dy}{dx} \) to test if it simplifies to one of the given answer choices.

Step 9: Simplification

After substituting, factor and simplify the expression using algebraic manipulation. Consider multiplicative constants and resemblance to the elements given in the multiple-choice answers.

Step 10: Identify Correct Answer

After simplification, recognize the form of the expression compared to \( n^2 y \). We find the expression simplifies to \( n^2 y \).

Key Concepts

Differentiation Rules

Differentiating functions is a foundational concept in calculus and is crucial in solving many mathematical problems. Differentiation rules help in finding the derivative of a given function. The three main rules commonly used are:

• Power Rule: If you have a function of the form \( f(x) = x^n \), the derivative \( f'(x) \) is \( n \cdot x^{n-1} \).
• Constant Rule: The derivative of a constant \( c \) is always zero, \( \frac{d}{dx}(c) = 0 \).
• Sum Rule: The derivative of a sum of functions is the sum of the derivatives of those functions, i.e., if \( f(x) = g(x) + h(x) \), then \( f'(x) = g'(x) + h'(x) \).

To apply these rules effectively, one must identify the structure of the function at hand. Differentiation becomes more involved when functions are nested, requiring more advanced techniques like the chain rule.

Chain Rule

The chain rule is a vital tool in differentiating composite functions, which are functions made up of other functions. If a function \( y \) is written as a composition, say \( y = f(g(x)) \), the chain rule tells us how to differentiate it:

• General Formula: \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
• This means we first differentiate the outer function \( f \) with respect to its inner function \( g \), then multiply this result by the derivative of the inner function \( g \).

In the given exercise, we see a clear application of the chain rule. The function \( y = (x + \sqrt{1+x^2})^n \) is tackled by treating the outer function as an exponential and the inner function \( x + \sqrt{1+x^2} \). This allows us to break down complex differentiation into manageable steps, adhering to the chain rule. Understanding when and how to use this rule is key to mastering implicit differentiation.

Product Rule

When two functions are multiplied together, their derivative isn't simply the product of their individual derivatives. Instead, we use the product rule. The product rule states:

• General Formula: If \( u(x) \) and \( v(x) \) are functions, \( \frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) \).

This principle helps in capturing the rate of change of a product. In the given solution, the product rule is essential when differentiating the expression for the second derivative \( \frac{d^2y}{dx^2} \). Here, the function is split into manageable components, and each part's derivative is calculated and combined systematically. Mastery of the product rule ensures that complex multiplications in functions can be differentiated while preserving all interactive effects between the components.