If y = sqrt{x^2 + 8x + 3}, then the value of | vedclass

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(A) Given $y = (x^2 + 8x + 3)^{1/2}$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{2}(x^2 + 8x + 3)^{-1/2} \cdot (2x + 8) = \frac{x +

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$\frac{-13}{216}$

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(A) Given $y = (x^2 + 8x + 3)^{1/2}$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{2}(x^2 + 8x + 3)^{-1/2} \cdot (2x + 8) = \frac{x + 4}{\sqrt{x^2 + 8x + 3}}$.Differentiating again using the quotient rule:$\frac{d^2y}{dx^2} = \frac{\sqrt{x^2 + 8x + 3} \cdot (1) - (x + 4) \cdot \frac{x + 4}{\sqrt{x^2 + 8x + 3}}}{x^2 + 8x + 3}$.$\frac{d^2y}{dx^2} = \frac{(x^2 + 8x + 3) - (x + 4)^2}{(x^2 + 8x + 3)^{3/2}} = \frac{x^2 + 8x + 3 - (x^2 + 8x + 16)}{(x^2 + 8x + 3)^{3/2}} = \frac{-13}{(x^2 + 8x + 3)^{3/2}}$.At $x = 3$, $x^2 + 8x + 3 = 9 + 24 + 3 = 36$.$\frac{d^2y}{dx^2} = \frac{-13}{(36)^{3/2}} = \frac{-13}{6^3} = \frac{-13}{216}$.

If $y = \sqrt{x^2 + 8x + 3}$, then the value of $\frac{d^2y}{dx^2}$ at $x = 3$ is

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