Question

If $x^2 + y^2 + \sin y = 4$, then the value of $\frac{d^2y}{dx^2}$ at $x=-2$ is

• A. -30
• B. -34
• C. -32
• D. -18

Hint:

Implicit differentiation. Carefully substitute the values of $x$, $y$, and $\frac{dy}{dx}$ into the second derivative expression.

Answer 1

The Correct Option is C

To solve the problem, we need to find the second derivative of \( y \) with respect to \( x \), denoted as \( \frac{d^2y}{dx^2} \), for the equation \( x^2 + y^2 + \sin y = 4 \) at \( x = -2 \).

First, differentiate the entire equation with respect to \( x \):

\(\frac{d}{dx}(x^2 + y^2 + \sin y) = \frac{d}{dx}(4)\)

Using the chain rule, we get:

\(2x + 2y\frac{dy}{dx} + \cos y\frac{dy}{dx} = 0\)

Rearrange to solve for \(\frac{dy}{dx}\):

\(2y\frac{dy}{dx} + \cos y\frac{dy}{dx} = -2x\)

\(\frac{dy}{dx}(2y + \cos y) = -2x\)

\(\frac{dy}{dx} = \frac{-2x}{2y + \cos y}\)

Next, we compute the second derivative. Differentiate \(\frac{dy}{dx}\) with respect to \( x \):

\(\frac{d}{dx}\left(\frac{-2x}{2y + \cos y}\right)\)

Using the quotient rule, \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2}\), let:

\(u = -2x\), \(v = 2y + \cos y\)

\(u' = -2\)

\(v' = 2\frac{dy}{dx} - \sin y\frac{dy}{dx}\)

Then,

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

Substitute \(\frac{dy}{dx} = \frac{-2x}{2y + \cos y}\) into this expression:

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

Now, find the values at \( x = -2 \). Plug in \( x = -2 \) into the original equation to solve for \( y \):

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

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

\(y^2 + \sin y = 0\)

If \( y = 0 \), then \(\sin y = 0\) and the equation holds: \(0^2 + \sin 0 = 0\).

Thus, \( y = 0 \) and \(\cos y = 1\).

Evaluate the derivatives:

\(\frac{dy}{dx} = \frac{-2(-2)}{2(0)+1} = \frac{4}{1} = 4\).

Substitute these into the second derivative expression:

\(\frac{d^2y}{dx^2} = \frac{(2(0)+1)(-2) - (-4)\left(2(4) - 0\right)}{(1)^2}\)

\(\frac{d^2y}{dx^2} = \frac{-2 - (-16)}{1}\)

\(\frac{d^2y}{dx^2} = -2 + 16 = -32\)

The value of \(\frac{d^2y}{dx^2}\) at \(x = -2\) is \(-32\).