If $y = (x + \sqrt{1+x^2})^n$, then $(1+x^2) \frac{d^2y}{dx^2} + x \frac{dy}{dx}$ is Options: (A) $n^2y$ (B) $-n^2y$ (C) $-y$ (D) $2x^2y$

### Analyzing the Setup The function provided is $y = (x + \sqrt{1+x^2})^n$. Our objective is to determine the value of the expression $(1+x^2) \frac{d^2y}{dx^2} + x \frac{dy}{dx}$. Rather than diving directly into a complex second derivative, we will use a strategic approach to simplify the expression. ### The First Derivative We begin by calculating the first derivative, $\frac{dy}{dx}$, using the chain rule: $$\frac{dy}{dx} = n(x + \sqrt{1+x^2})^{n-1} \cdot \left( 1 + \frac{2x}{2\sqrt{1+x^2}} \right)$$ Simplifying the term inside the parenthesis, we obtain: $$\frac{dy}{dx} = n(x + \sqrt{1+x^2})^{n-1} \cdot \left( \frac{\sqrt{1+x^2} + x}{\sqrt{1+x^2}} \right)$$ Since the base $(x + \sqrt{1+x^2})$ is common, we add the exponents $(n-1) + 1 = n$. This yields the elegant relation: $$\frac{dy}{dx} = \frac{ny}{\sqrt{1+x^2}}$$ ### The Strategic Pivot To avoid the messy quotient rule, we cross-multiply to isolate the radical: $$\sqrt{1+x^2} \frac{dy}{dx} = ny$$ To eliminate the square root entirely, we square both sides of the equation: $$(1+x^2) \left( \frac{dy}{dx} \right)^2 = n^2 y^2$$ ### The Final Act Now, we differentiate both sides with respect to $x$ using the product rule on the left and the chain rule on the right: $$(1+x^2) \cdot 2 \left( \frac{dy}{dx} \right) \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 \cdot 2x = n^2 \cdot 2y \frac{dy}{dx}$$ We observe that $2 \frac{dy}{dx}$ is a common factor across all terms. Dividing the entire equation by $2 \frac{dy}{dx}$ (assuming $\frac{dy}{dx} \neq 0$), we arrive at the final result: $$(1+x^2) \frac{d^2y}{dx^2} + x \frac{dy}{dx} = n^2 y$$ The final value of the expression is **$n^2 y$**.

JEE Main 2002

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: If $y = (x + 1 + x^{2})^{n}$ , then $(1 + x^{2}) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x}$ is

Select Answer:

Progressive Hints

Hint 1

The function

y = (x + 1 + x^{2})^{n}

looks intimidating, but don't rush to differentiate it directly. Notice that if you differentiate

y

, you will likely get a term involving

1 + x^{2}

in the denominator. Try to express the first derivative

\frac{d y}{d x}

in terms of

y

itself.

Hint 2

Instead of using the quotient rule for the second derivative, which will lead to a mess of algebraic terms, try cross-multiplying the

1 + x^{2}

term to the left side. This transforms the equation into a product of two functions, making the subsequent differentiation much cleaner.

Hint 3

Before you differentiate for the second time, square both sides of your equation

1 + x^{2} \frac{d y}{d x} = n y

. This will eliminate the square root entirely, leaving you with a simple polynomial-like structure that is easy to differentiate using the product rule.

Visualized Solution

Identify the Function $y$

Given function: $y = (x + 1 + x^{2})^{n}$
Goal: Find the value of $(1 + x^{2}) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x}$
This requires finding the first and second derivatives.

First Differentiation (Chain Rule)

Differentiating $y$ with respect to $x$ :
$\frac{d y}{d x} = \frac{d}{d x} [(x + 1 + x^{2})^{n}]$
We must apply the Chain Rule: $\frac{d}{d x} [u^{n}] = n u^{n - 1} \cdot \frac{d u}{d x}$

Applying the Chain Rule

$\frac{d y}{d x} = n (x + 1 + x^{2})^{n - 1} \cdot \frac{d}{d x} (x + 1 + x^{2})$
Now, differentiate the inner term.

Differentiating the Inner Term

Derivative of $x$ is $1$ .
Derivative of $1 + x^{2}$ is $\frac{1}{2 1 + x ^{2}} \cdot 2 x$ .
$\frac{d y}{d x} = n (x + 1 + x^{2})^{n - 1} \cdot (1 + \frac{2 x}{2 1 + x ^{2}})$

Simplify the Inner Derivative

Cancel the $2$ in the numerator and denominator.
$\frac{d y}{d x} = n (x + 1 + x^{2})^{n - 1} \cdot (1 + \frac{x}{1 + x ^{2}})$
Take the LCM inside the bracket:
$\frac{d y}{d x} = n (x + 1 + x^{2})^{n - 1} \cdot (\frac{1 + x ^{2} + x}{1 + x ^{2}})$

Combine the Terms

Notice that $(x + 1 + x^{2})^{n - 1} \cdot (x + 1 + x^{2}) = (x + 1 + x^{2})^{n}$
$\frac{d y}{d x} = \frac{n ( x + 1 + x ^{2} ) ^{n}}{1 + x ^{2}}$

Express $y^{'}$ in terms of $y$

Recall the original function: $y = (x + 1 + x^{2})^{n}$
Substitute $y$ back into the derivative:
$\frac{d y}{d x} = \frac{n y}{1 + x ^{2}}$

Rearrange to Avoid Quotient Rule

$1 + x^{2} \frac{d y}{d x} = n y$
To find the second derivative, we could use the quotient rule, but it is messy.
Pro Tip: Cross-multiply to convert it into a product rule problem.

Square Both Sides

Square both sides to eliminate the square root completely:
$(1 + x^{2}) (\frac{d y}{d x})^{2} = n^{2} y^{2}$

Second Differentiation Setup

Differentiate both sides with respect to $x$ :
$\frac{d}{d x} [(1 + x^{2}) (\frac{d y}{d x})^{2}] = \frac{d}{d x} [n^{2} y^{2}]$
We will use the Product Rule on the left: $(uv)^{'} = u^{'} v + u v^{'}$
And the Chain Rule on the right.

Applying Product and Chain Rules

Left side (Product Rule):
$(1 + x^{2}) \cdot \frac{d}{d x} [(\frac{d y}{d x})^{2}] + (\frac{d y}{d x})^{2} \cdot \frac{d}{d x} [1 + x^{2}]$
Right side (Chain Rule):
$n^{2} \cdot \frac{d}{d x} [y^{2}]$

Executing the Differentiation

$\frac{d}{d x} [(\frac{d y}{d x})^{2}] = 2 (\frac{d y}{d x}) \cdot \frac{d ^{2} y}{d x ^{2}}$
$\frac{d}{d x} [1 + x^{2}] = 2 x$
$\frac{d}{d x} [y^{2}] = 2 y \cdot \frac{d y}{d x}$
Equation becomes:
$(1 + x^{2}) \cdot 2 (\frac{d y}{d x}) \frac{d ^{2} y}{d x ^{2}} + (\frac{d y}{d x})^{2} \cdot 2 x = n^{2} \cdot 2 y \frac{d y}{d x}$

Final Simplification

Notice that $2 \frac{d y}{d x}$ is a common factor in every term.
Divide the entire equation by $2 \frac{d y}{d x}$ (assuming $\frac{d y}{d x} \neq = 0$ ):
$(1 + x^{2}) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x} = n^{2} y$

Conclusion and Key Takeaway

The value of $(1 + x^{2}) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x}$ is $n^{2} y$ .
Key Takeaway: When dealing with complex powers, always try to express the first derivative in terms of the original function $y$ before finding the second derivative.

00:00 / 00:00

The Sigma Insight: Higher Order Derivatives

Solution Diagram

Analyzing the Setup

The function provided is

y = (x + 1 + x^{2})^{n}

. Our objective is to determine the value of the expression

(1 + x^{2}) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x}

Rather than diving directly into a complex second derivative, we will use a strategic approach to simplify the expression.

The First Derivative

We begin by calculating the first derivative,

\frac{d y}{d x}

, using the chain rule:

\frac{d y}{d x} = n (x + 1 + x^{2})^{n - 1} \cdot (1 + \frac{2 x}{2 1 + x ^{2}})

Simplifying the term inside the parenthesis, we obtain:

\frac{d y}{d x} = n (x + 1 + x^{2})^{n - 1} \cdot (\frac{1 + x ^{2} + x}{1 + x ^{2}})

Since the base

(x + 1 + x^{2})

is common, we add the exponents

(n - 1) + 1 = n

. This yields the elegant relation:

\frac{d y}{d x} = \frac{n y}{1 + x ^{2}}

The Strategic Pivot

To avoid the messy quotient rule, we cross-multiply to isolate the radical:

1 + x^{2} \frac{d y}{d x} = n y

To eliminate the square root entirely, we square both sides of the equation:

(1 + x^{2}) (\frac{d y}{d x})^{2} = n^{2} y^{2}

The Final Act

Now, we differentiate both sides with respect to

x

using the product rule on the left and the chain rule on the right:

(1 + x^{2}) \cdot 2 (\frac{d y}{d x}) \frac{d ^{2} y}{d x ^{2}} + (\frac{d y}{d x})^{2} \cdot 2 x = n^{2} \cdot 2 y \frac{d y}{d x}

We observe that

2 \frac{d y}{d x}

is a common factor across all terms. Dividing the entire equation by

2 \frac{d y}{d x}

(assuming $\frac{dy}{dx} eq 0$), we arrive at the final result:

(1 + x^{2}) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x} = n^{2} y

The final value of the expression is $n^{2} y$ .

Common Traps & FAQs

Wait, why do we reject the quotient rule here? Isn't it just a standard method?

You are absolutely right that the quotient rule is a standard method. However, in JEE Advanced, time and accuracy are your best friends. Using the quotient rule here would introduce a square root in the denominator, which would then require the quotient rule again for the second derivative, leading to a very high probability of a sign error or a messy algebraic mistake. The cross-multiplication trick is a strategic choice to keep the algebra elegant and error-free.

Is it always valid to divide by

2 \frac{d y}{d x}

in the final step?

That is a sharp observation! Mathematically, we can only divide by an expression if it is non-zero. In this specific case,

\frac{d y}{d x} = n (x + 1 + x^{2})^{n - 1} (1 + \frac{x}{1 + x ^{2}})

. Since

x + 1 + x^{2}

is never zero for real

x

, and the term in the bracket is also non-zero,

\frac{d y}{d x}

is non-zero. Thus, the division is perfectly valid.

How did you know to square both sides? That feels like a random step.

It is not random at all! It is a classic technique used when dealing with differential equations involving square roots. By squaring, we remove the radical, which simplifies the structure of the equation significantly. Whenever you see a square root in a derivative expression that you need to differentiate again, squaring is almost always the path to the simplest solution.

Animated Solution for Mathematics - Differentiation: If $y = (x + 1 + x^{2})^{n}$ , then $(1 + x^{2}) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x}$ is

Select Answer:

Visualized Solution

Identify the Function $y$

First Differentiation (Chain Rule)

Applying the Chain Rule

Differentiating the Inner Term

Simplify the Inner Derivative

Combine the Terms

Express $y^{'}$ in terms of $y$

Rearrange to Avoid Quotient Rule

Square Both Sides

Second Differentiation Setup

Applying Product and Chain Rules

Executing the Differentiation

Final Simplification

Conclusion and Key Takeaway

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

The First Derivative

The Strategic Pivot

The Final Act

Similar Questions

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If $x^{2} + y^{2} = 1$ , then

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If $(a + b x) e^{y / x} = x$ , then prove that $x^{3} \frac{d ^{2} y}{d x ^{2}} = (x \frac{d y}{d x} - y)^{2}$ .

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=(x+1+x2​)n, then (1+x2)dx2d2y​+xdxdy​ is

Select Answer:

Visualized Solution

Identify the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

First Differentiation (Chain Rule)

Applying the Chain Rule

Differentiating the Inner Term

Simplify the Inner Derivative

Combine the Terms

Express .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y′ in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Rearrange to Avoid Quotient Rule

Square Both Sides

Second Differentiation Setup

Applying Product and Chain Rules

Executing the Differentiation

Final Simplification

Conclusion and Key Takeaway

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

The First Derivative

The Strategic Pivot

The Final Act

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Animated Solution for Mathematics - Differentiation: If $y = (x + 1 + x^{2})^{n}$ , then $(1 + x^{2}) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x}$ is

Identify the Function $y$

Express $y^{'}$ in terms of $y$

$\frac{d ^{2} x}{d y ^{2}}$ equals

If $y^{2} = P (x)$ , a polynomial of degree 3, then $2 \frac{d}{d x} (y^{3} \frac{d ^{2} y}{d x ^{2}})$ equals

$\frac{d ^{2} x}{d y ^{2}}$ equals

If $x^{2} + y^{2} = 1$ , then

If $x^{2} + y^{2} = 1$ then

If $(a + b x) e^{y / x} = x$ , then prove that $x^{3} \frac{d ^{2} y}{d x ^{2}} = (x \frac{d y}{d x} - y)^{2}$ .

If $lo g_{e} y = 3 sin^{- 1} x$ , then $(1 - x^{2}) y^{''} - x y^{'}$ at $x = \frac{1}{2}$ is equal to

If $y (x) = (x^{x^{x}}), x > 0$ then $\frac{d ^{2} x}{d y ^{2}} + 20$ at $x = 1$ is equal to:

If $y = tan^{- 1} (sec x^{3} - tan x^{3})$ , $\frac{π}{2} < x^{3} < \frac{3 π}{2}$ , then

Let $y = lo g_{e} (\frac{1 - x ^{2}}{1 + x ^{2}}), - 1 < x < 1$ . Then at $x = \frac{1}{2}$ , the value of $225 (y^{'} - y^{''})$ is equal to