IGDTUW 2025 — Mathematics PYQ

Step 1: Find the first derivative $\frac{dy}{dx}$

Given:

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

Differentiating with respect to $x$ using the chain rule:

$$\frac{dy}{dx}=n(x+\sqrt{1+x^2}{)}^{n-1}\cdot \frac{d}{dx}(x+\sqrt{1+x^2})$$

$$\frac{dy}{dx}=n(x+\sqrt{1+x^2}{)}^{n-1}\cdot \left(1+\frac{1}{2\sqrt{1+x^2}}\cdot 2x\right)$$

$$\frac{dy}{dx}=n(x+\sqrt{1+x^2}{)}^{n-1}\cdot \left(\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}\right)$$

Since $(x+\sqrt{1+x^2}{)}^{n-1}\cdot (x+\sqrt{1+x^2})=(x+\sqrt{1+x^2}{)}^n=y$, we have:

$$\frac{dy}{dx}=\frac{ny}{\sqrt{1+x^2}}$$

$$\sqrt{1+x^2}\frac{dy}{dx}=ny$$

Step 2: Find the second derivative $\frac{d^{2}y}{d{x}^2}$

Differentiate both sides with respect to $x$ using the product rule on the left side:

$$\frac{d}{dx}\left(\sqrt{1+x^2}\frac{dy}{dx}\right)=\frac{d}{dx}(ny)$$

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

$$\sqrt{1+x^2}\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}\cdot \left(\frac{x}{\sqrt{1+x^2}}\right)=n\frac{dy}{dx}$$

Multiply the entire equation by $\sqrt{1+x^2}$ to clear the denominator:

$$(1+x^2)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=n\sqrt{1+x^2}\frac{dy}{dx}$$

Step 3: Substitute the value of $\frac{dy}{dx}$

From Step 1, we know that $\sqrt{1+x^2}\frac{dy}{dx}=ny$. Substitute this into the right side of the equation:

$$(1+x^2)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=n(ny)$$

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

Conclusion:

The expression is equal to $n^{2}y$. Therefore, the correct option is (a).