If y = (x + sqrt{1 + x^2})^n,then (1 + x^2)fr | vedclass

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(A) Given $y = (x + \sqrt{1 + x^2})^n$. Differentiating with respect to $x$,we get: $\frac{dy}{dx} = n(x + \sqrt{1 + x^2})^{n-1} \cdot (1 + \frac{x}{\

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$n^2y$

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(A) Given $y = (x + \sqrt{1 + x^2})^n$. Differentiating with respect to $x$,we get: $\frac{dy}{dx} = n(x + \sqrt{1 + x^2})^{n-1} \cdot (1 + \frac{x}{\sqrt{1 + x^2}})$ $\frac{dy}{dx} = n(x + \sqrt{1 + x^2})^{n-1} \cdot \frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2}}$ $\frac{dy}{dx} = \frac{n(x + \sqrt{1 + x^2})^n}{\sqrt{1 + x^2}}$ $\sqrt{1 + x^2} \frac{dy}{dx} = ny$. Differentiating both sides with respect to $x$ again: $\frac{d}{dx}(\sqrt{1 + x^2} \frac{dy}{dx}) = \frac{d}{dx}(ny)$ $\sqrt{1 + x^2} \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{x}{\sqrt{1 + x^2}} = n \frac{dy}{dx}$ Multiply both sides by $\sqrt{1 + x^2}$: $(1 + x^2) \frac{d^2y}{dx^2} + x \frac{dy}{dx} = n \sqrt{1 + x^2} \frac{dy}{dx}$ Since $\sqrt{1 + x^2} \frac{dy}{dx} = ny$,we substitute this into the equation: $(1 + x^2) \frac{d^2y}{dx^2} + x \frac{dy}{dx} = n(ny) = n^2y$.

If $y = (x + \sqrt{1 + x^2})^n$,then $(1 + x^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx}$ is equal to:

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