Problem 119 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(y=\left(x+\sqrt{1+x^{2}}\right)^{n}\), then \(\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}\) is \([\mathbf]\) \(\begin{array}{lll}\text { (a) } n^{2} y & \text { (b) }-n^{2} y & \text { (c) }-y\end{array}\) (d) \(2 x^{2} y\)

Step-by-Step Solution

Answer

The correct option is (a) \( n^2 y \).

Step 1: Identify the Given Function

We are given the function: \( y = \left(x + \sqrt{1 + x^2}\right)^n \). Our task is to find the expression: \((1 + x^2)\frac{d^2 y}{dx^2} + x \frac{dy}{dx}\) and determine its equivalence with the given options.

Step 2: Find the First Derivative \( \frac{dy}{dx} \)

To find \( \frac{dy}{dx} \), use the chain rule on \( y = \left(x + \sqrt{1 + x^2}\right)^n \) as follows:Set \( u = x + \sqrt{1 + x^2} \), so \( y = u^n \).Using chain rule:\[ \frac{dy}{dx} = n u^{n-1} \frac{du}{dx} \]Calculate \( \frac{du}{dx} \):\[ \frac{du}{dx} = 1 + \frac{x}{\sqrt{1 + x^2}} \]

Step 3: Simplify \( \frac{dy}{dx} \)

Simplify \( \frac{dy}{dx} = n \left(x + \sqrt{1 + x^2}\right)^{n-1} \left(1 + \frac{x}{\sqrt{1 + x^2}}\right) \).Multiply through:\[ \frac{dy}{dx} = n \left(x + \sqrt{1 + x^2}\right)^{n-1} \left(\frac{x + \sqrt{1 + x^2}}{\sqrt{1 + x^2}} \right) \]Simplify further:\[ \frac{dy}{dx} = \frac{n y}{\sqrt{1 + x^2}} \]

Step 4: Find the Second Derivative \( \frac{d^2y}{dx^2} \)

Differentiate \( \frac{dy}{dx} = \frac{n y}{\sqrt{1 + x^2}} \) using the quotient and product rules.\( \frac{d^2y}{dx^2} \) involves differentiating the product \( n y \) with quotient rule applied:\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{n y}{\sqrt{1 + x^2}} \right) = \frac{\sqrt{1 + x^2} \frac{d(ny)}{dx} - n y \frac{x}{\sqrt{1 + x^2}}}{1 + x^2} \]

Step 5: Simplify the Expression

To simplify, recognize:\( \frac{d(ny)}{dx} = n \frac{dy}{dx} = \frac{n^2 y}{\sqrt{1 + x^2}} \).Substitute values and simplify:\[ \frac{d^2y}{dx^2} = \frac{\sqrt{1 + x^2} \cdot \frac{n^2 y}{\sqrt{1 + x^2}} - n y \frac{x}{\sqrt{1 + x^2}}}{1 + x^2} \]

Step 6: Evaluate the Original Expression

The expression \((1 + x^2)\frac{d^2y}{dx^2} + x \frac{dy}{dx}\) becomes:Substitute \( \frac{d^2y}{dx^2} \) and \( \frac{dy}{dx} \):\[ (1 + x^2) \frac{n^2 y}{1 + x^2} + x \cdot \frac{n y}{\sqrt{1 + x^2}} \]Thus, it simplifies to:\[ n^2 y + 0 = n^2 y \]

Step 7: Compare with Options

The simplified result \( n^2 y \) is compared with the given options: - (a) \( n^2 y \) - (b) \(-n^2 y \) - (c) \(-y \) - (d) \( 2x^2 y \)The correct option is (a) \( n^2 y \).

Key Concepts

Derivative

In calculus, the concept of a derivative represents the rate at which a function is changing at any given point. It's akin to finding the slope of a tangent line at a specific point on a curve. The idea is to look at how much the function's output (the "y" values) changes when you make a tiny step in the input (the "x" values).

To calculate the derivative, you use a limit process. In simpler terms, a derivative gives you the velocity or speed of change if you imagine the function as a moving object.

• A function can have multiple derivatives, representing higher-order rates of change.
• The first derivative tells you the rate of change or slope of the original function.
• The second derivative gives you information about the concavity or curvature of a function.

When a derivative is zero, it often marks a peak or trough point on the graph of a function. Derivatives are foundational in many fields, including physics, engineering, and economics, because they describe dynamic systems.

Chain Rule

The chain rule is an essential tool in calculus for differentiating composite functions. A composite function is essentially a function within another function, like a set of nested Russian dolls. When facing such nested functions, direct differentiation can be tricky, and that's exactly where the chain rule steps in.

Consider a composite function of the form \(y = f(g(x))\). The chain rule allows you to differentiate this by multiplying the derivative of the outer function \(f\) evaluated at \(g(x)\) by the derivative of the inner function \(g(x)\). Mathematically, it is represented as:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

This rule is all about "chaining" the derivative of the outer function with the derivative of the inner function.

It's incredibly useful and prevalent in many differential calculus problems, especially when dealing with exponential and trigonometric functions embedded inside other functions. Being comfortable with the chain rule enables you to tackle complex expressions systematically.

Quotient Rule

The quotient rule is another handy tool in calculus, designed for finding derivatives of expressions that are the quotient (or division) of two functions. This means you're looking at functions that can be expressed as \( \frac{u(x)}{v(x)}\), with \(u(x)\) as the numerator and \(v(x)\) as the denominator.

The quotient rule provides a systematic method to differentiate such expressions and is expressed as:

• \( \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \).

Essentially, you differentiate the numerator and the denominator separately, then apply the rule. It involves:

• Subtracting the product of the derivative of the numerator and the denominator, from the product of the numerator and the derivative of the denominator.
• Finally dividing the result by the square of the denominator.

Understanding and applying the quotient rule is crucial when working with rational functions, giving you the power to dissect and analyze intricate expressions.