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(B) Given $x = \int\limits_0^y \frac{dt}{\sqrt{1 + t^2}}$.By the Fundamental Theorem of Calculus,differentiating both sides with respect to $x$ gives:

Vedclass · Accepted Answer

$y$

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(B) Given $x = \int\limits_0^y \frac{dt}{\sqrt{1 + t^2}}$.By the Fundamental Theorem of Calculus,differentiating both sides with respect to $x$ gives:$\frac{dx}{dx} = \frac{d}{dx} \left( \int\limits_0^y \frac{dt}{\sqrt{1 + t^2}} \right)$$1 = \frac{1}{\sqrt{1 + y^2}} \cdot \frac{dy}{dx}$Therefore,$\frac{dy}{dx} = \sqrt{1 + y^2}$.Now,differentiate $\frac{dy}{dx}$ with respect to $x$ to find $\frac{d^2y}{dx^2}$:$\frac{d^2y}{dx^2} = \frac{d}{dx} (\sqrt{1 + y^2})$$\frac{d^2y}{dx^2} = \frac{1}{2\sqrt{1 + y^2}} \cdot 2y \cdot \frac{dy}{dx}$Substitute $\frac{dy}{dx} = \sqrt{1 + y^2}$ into the expression:$\frac{d^2y}{dx^2} = \frac{y}{\sqrt{1 + y^2}} \cdot \sqrt{1 + y^2} = y$.Thus,the correct option is $B$.

Variables $x$ and $y$ are related by the equation $x = \int\limits_0^y \frac{dt}{\sqrt{1 + t^2}}$. The value of $\frac{d^2y}{dx^2}$ is equal to

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