If `2 sin x. cos y=1, ` then `(d ^(2)y)/(dx ^(2)) at ((pi)/(4), (pi)/(4))` is …….

Options

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

Answer

To find \(\frac{d^2y}{dx^2}\) at the point \(\left(\frac{\pi}{4}, \frac{\pi}{4}\right)\) given the equation \(2 \sin x \cos y = 1\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 2 \sin x \cos y = 1 \]
Dividing both sides by 2, we get:
\[ \sin x \cos y = \frac{1}{2} \]

### Step 2: Differentiate with respect to \(x\)
Now, we differentiate 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:
\[ \cos x \cos y + \sin x \left(-\sin y \frac{dy}{dx}\right) = 0 \]
This simplifies to:
\[ \cos x \cos y - \sin x \sin y \frac{dy}{dx} = 0 \]

### Step 3: Solve for \(\frac{dy}{dx}\)
Rearranging the equation gives:
\[ \sin x \sin y \frac{dy}{dx} = \cos x \cos y \]
Thus, we find:
\[ \frac{dy}{dx} = \frac{\cos x \cos y}{\sin x \sin y} \]

### Step 4: Differentiate \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2}\)
Now we differentiate \(\frac{dy}{dx}\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{\cos x \cos y}{\sin x \sin y}\right) \]
Using the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{(\sin x \sin y)(-\sin x \cos y + \cos x \sin y \frac{dy}{dx}) - (\cos x \cos y)(\cos x \sin y + \sin x \cos y \frac{dy}{dx})}{(\sin x \sin y)^2} \]

### Step 5: Substitute \(x = \frac{\pi}{4}\) and \(y = \frac{\pi}{4}\)
At \(x = \frac{\pi}{4}\) and \(y = \frac{\pi}{4}\):
- \(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)
- \(\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)

Substituting these values into the expression for \(\frac{d^2y}{dx^2}\):
\[ \frac{d^2y}{dx^2} = \frac{\left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}\right)\left(-\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} \cdot \frac{dy}{dx}\right) - \left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} \cdot \frac{dy}{dx}\right)}{\left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}\right)^2} \]

### Step 6: Simplify the expression
After substituting and simplifying, we find that:
\[ \frac{d^2y}{dx^2} = -4 \]

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