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Differentiation

Section: Mathematics

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Question : 1 of 34

Marks: +1, -0

If

x^2 + y^2 + \sin \; y = 4

, then the value of

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

x = -2

[AP EAPCET 24 May 2025 S1]

$-30$
$-34$
$-32$
$-18$

Solution:

Given,

x^2 + y^2 + \sin \; y = 4

Differentiate both sides w.r.t

x

, we get

\; 2 x + 2 y \cdot \; \frac{d y}{d x} + \cos y \cdot \; \frac{d y}{d x} = 0 \Rightarrow \; \frac{d y}{d x} (2 y + \cos y) = -2 x

\; \; \frac{d y}{d x} = \; \frac{-2 x}{2 y + \cos y}

Again, differentiate both sides w.r.t

x

, we get

\; \frac{d^2 y}{d x^2} \; = \; \frac{d}{d x} \left( \; \frac{-2 x}{2 y + \cos y} \right)

\; = \; \frac{(2 y + \cos y) \cdot (-2) - (-2 x) \cdot (2 - \sin \; y) \cdot \frac{d y}{d x}}{(2 y + \cos y)^2}

\; = \; \frac{-4 y - 2 \cos y + 2 x \cdot (2 - \sin \; y) \cdot \left( \frac{-2 x}{2 y + \cos y} \right)}{(2 y + \cos y)^2}

\; = \; \frac{-4 y - 2 \cos y - \frac{4 x^2 (2 - \sin \; y)}{2 y + \cos y}}{(2 y + \cos y)^2}

Put

x = -2

in original equation, we get

\; (-2)^2 + y^2 + \sin \; y = 4

\Rightarrow \; \; y^2 + \sin \; y = 0 \Rightarrow y = 0

Now, from Eq. (i), we get

\; \frac{d^2 y}{d x^2} \; = \; \frac{-4(0) - 2 \cos(0) - \frac{4(-2)^2 (2 - \sin(0))}{2 \cdot (0) + \cos(0)}}{(2(0) + \cos(0))^2}

\; = \; \frac{-2 - \frac{16(2-0)}{1}}{1^2} = \; \frac{-2-32}{1} = -34

\therefore \; \; \frac{d^2 y}{d x^2} \; \frac{}{} |_{x=-2} \; = -34

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