If y={x²+8 x+3}, then the value of {d² y}{d x²} at x=3 is

Correct Answer is : −13216-13/216216−13

Correct Answer is : −13216-13/216216−13

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MHTCET 11 Apr 2026 Shift 1 Paper

Question : 120 of 150

Marks: +1, -0

y=\sqrt{x^{2}+8 x+3}

\frac{d^{2} y}{d x^{2}}

x=3

Solution:

Concept:

Differentiate the given function twice using the chain rule and the product rule, then substitute

x = 3

Explanation:

Given:

y = \sqrt{x^2 + 8x + 3} = \left(x^2 + 8x + 3\right)^{\frac{1}{2}}

Differentiate once with respect to

x

\frac{dy}{dx} = \frac{1}{2}\left(x^2 + 8x + 3\right)^{-\frac{1}{2}}(2x + 8)

Simplify:

\frac{dy}{dx} = \frac{x + 4}{\sqrt{x^2 + 8x + 3}}

Differentiate again using the product rule:

\frac{d^2y}{dx^2} = \left(x^2 + 8x + 3\right)^{-\frac{1}{2}} + (x + 4)\left[-\frac{1}{2}\left(x^2 + 8x + 3\right)^{-\frac{3}{2}}(2x + 8)\right]

Since

2x + 8 = 2(x + 4)

\frac{d^2y}{dx^2} = \frac{1}{\sqrt{x^2 + 8x + 3}} - \frac{(x + 4)^2}{\left(x^2 + 8x + 3\right)^{\frac{3}{2}}}

Taking a common denominator:

\frac{d^2y}{dx^2} = \frac{x^2 + 8x + 3 - (x + 4)^2}{\left(x^2 + 8x + 3\right)^{\frac{3}{2}}}

Now,

(x + 4)^2 = x^2 + 8x + 16

x^2 + 8x + 3 - \left(x^2 + 8x + 16\right) = -13

Thus:

\frac{d^2y}{dx^2} = \frac{-13}{\left(x^2 + 8x + 3\right)^{\frac{3}{2}}}

x = 3

x^2 + 8x + 3 = 9 + 24 + 3 = 36

So:

\frac{d^2y}{dx^2}\bigg|_{x=3} = \frac{-13}{36^{\frac{3}{2}}}

Since

36^{\frac{3}{2}} = (\sqrt{36})^3 = 6^3 = 216

\frac{d^2y}{dx^2}\bigg|_{x=3} = \frac{-13}{216}

Answer:

The value of

\frac{d^2y}{dx^2}

x = 3

\frac{-13}{216}

Correct option: A.

\frac{-13}{216}

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