Question:

Step 1: Set up the given function and find the first derivative. We are given $y = (x + \sqrt{1+x^2})^n$. To find $\frac{dy}{dx}$, we apply the chain rule: $$\frac{dy}{dx} = n(x+\sqrt{1+x^2})^{n-1} \cdot \frac{d}{dx}(x+\sqrt{1+x^2})$$ Computing the derivative of the inner function: $$\frac{d}{dx}(x+\sqrt{1+x^2}) = 1 + \frac{x}{\sqrt{1+x^2}} = \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}$$ Therefore: $$\frac{dy}{dx} = n(x+\sqrt{1+x^2})^{n-1} \cdot \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}$$ Step 2: Simplify the first derivative. Notice that $(x+\sqrt{1+x^2})^{n-1} \cdot (x+\sqrt{1+x^2}) = (x+\sqrt{1+x^2})^n = y$. Thus: $$\frac{dy}{dx} = \frac{n(x+\sqrt{1+x^2})^n}{\sqrt{1+x^2}} = \frac{ny}{\sqrt{1+x^2}}$$ This gives us the key relation: $$\sqrt{1+x^2} \cdot \frac{dy}{dx} = ny$$ Squaring both sides: $$(1+x^2)\left(\frac{dy}{dx}\right)^2 = n^2 y^2$$ Step 3: Differentiate the squared relation with respect to $x$. Differentiating $(1+x^2)(y')^2 = n^2 y^2$ with respect to $x$ (where $y' = \frac{dy}{dx}$ and $y'' = \frac{d^2y}{dx^2}$): $$\frac{d}{dx}[(1+x^2)(y')^2] = \frac{d}{dx}[n^2 y^2]$$ Using the product rule on the left side: $$2x(y')^2 + (1+x^2) \cdot 2y' \cdot y'' = 2n^2 y \cdot y'$$ Step 4: Simplify by dividing by $2y'$. Assuming $y' \neq 0$, we divide the entire equation by $2y'$: $$x(y') + (1+x^2)y'' = n^2 y$$ Rearranging: $$(1+x^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx} = n^2 y$$ Step 5: State the final answer. The value of $(1+x^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx}$ is equal to $n^2 y$. The correct answer is **Option 3: $n^2 y$**.