If $x^2 + y^2 + \sin y = 4$, then the value of $\frac{d^2y}{dx^2}$ at $x=-2$ is
Show Hint
- -30
- -34
- -32
- -18
The Correct Option is C
Solution and Explanation
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 \(-32\).
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