If `(sinx)(cosy)=1/2,` then `(d^2y)/(dx^2)` at `(pi/4,pi/4)` is

Options

A. `-4`
B. `-2`
C. `-6`
D. `0`

Answer

To solve the problem, we need to find the second derivative \(\frac{d^2y}{dx^2}\) given the equation \((\sin x)(\cos y) = \frac{1}{2}\) at the point \((\frac{\pi}{4}, \frac{\pi}{4})\).

### Step 1: Differentiate the given equation
We start with the equation:
\[ \sin x \cos y = \frac{1}{2} \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}(\sin x \cos y) = \frac{d}{dx}\left(\frac{1}{2}\right) \]
Using the product rule on the left side:
\[ \cos x \cos y + \sin x \left(-\sin y \frac{dy}{dx}\right) = 0 \]

### Step 2: Rearranging the equation
Rearranging the differentiated equation gives:
\[ \cos x \cos y = \sin x \sin y \frac{dy}{dx} \]
Thus,
\[ \frac{dy}{dx} = \frac{\cos x \cos y}{\sin x \sin y} \]
This can be simplified to:
\[ \frac{dy}{dx} = \cot x \cot y \]

### Step 3: Differentiate again to find \(\frac{d^2y}{dx^2}\)
Now we differentiate \(\frac{dy}{dx} = \cot x \cot y\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(\cot x \cot y) \]
Using the product rule again:
\[ \frac{d^2y}{dx^2} = -\csc^2 x \cot y + \cot x (-\csc^2 y \frac{dy}{dx}) \]
Substituting \(\frac{dy}{dx} = \cot x \cot y\):
\[ \frac{d^2y}{dx^2} = -\csc^2 x \cot y - \cot x \csc^2 y \cot x \cot y \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = -\csc^2 x \cot y - \cot^2 x \csc^2 y \cot y \]

### Step 4: Evaluate at \(x = \frac{\pi}{4}, y = \frac{\pi}{4}\)
Now we substitute \(x = \frac{\pi}{4}\) and \(y = \frac{\pi}{4}\):
- \(\cot\left(\frac{\pi}{4}\right) = 1\)
- \(\csc^2\left(\frac{\pi}{4}\right) = 2\)

Substituting these values into the equation:
\[ \frac{d^2y}{dx^2} = -2(1) - (1^2)(2)(1) = -2 - 2 = -4 \]

### Final Answer
Thus, the value of \(\frac{d^2y}{dx^2}\) at \((\frac{\pi}{4}, \frac{\pi}{4})\) is:
\[ \frac{d^2y}{dx^2} = -4 \]