If x = intlimits_0^y {frac{{dt}}{{sqrt {1 + { | vedclass

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(A) Given $x = \int\limits_0^y {\frac{{dt}}{{\sqrt {1 + {t^2}} }}} $. Applying the Leibniz rule for differentiation under the integral sign,we differe

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$y$

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(A) Given $x = \int\limits_0^y {\frac{{dt}}{{\sqrt {1 + {t^2}} }}} $. Applying the Leibniz rule for differentiation under the integral sign,we differentiate both sides with respect to $x$: $\frac{dx}{dx} = \frac{1}{\sqrt{1 + y^2}} \cdot \frac{dy}{dx}$. $1 = \frac{1}{\sqrt{1 + y^2}} \cdot \frac{dy}{dx}$. $\frac{dy}{dx} = \sqrt{1 + y^2}$. Now,differentiate both sides with respect to $x$ again to find $\frac{d^2y}{dx^2}$: $\frac{d^2y}{dx^2} = \frac{d}{dx}(\sqrt{1 + y^2}) = \frac{1}{2\sqrt{1 + y^2}} \cdot \frac{d}{dx}(1 + y^2)$. $\frac{d^2y}{dx^2} = \frac{1}{2\sqrt{1 + y^2}} \cdot (2y \cdot \frac{dy}{dx})$. Substitute $\frac{dy}{dx} = \sqrt{1 + y^2}$ into the equation: $\frac{d^2y}{dx^2} = \frac{y}{\sqrt{1 + y^2}} \cdot \sqrt{1 + y^2} = y$. Thus,$\frac{d^2y}{dx^2} = y$.

If $x = \int\limits_0^y {\frac{{dt}}{{\sqrt {1 + {t^2}} }}} $,then $\frac{{{d^2}y}}{{d{x^2}}}$ is equal to

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