If `xe^(xy)-y=sin^(2)x` then `(dy)/(dx)` at x = 0 is

Options

A. 0
B. 1
C. `-1`
D. none of these

Explanation

To find \(\frac{dy}{dx}\) at \(x = 0\) for the equation \(xe^{xy} - y = \sin^2 x\), we will follow these steps:

### Step 1: Differentiate both sides with respect to \(x\)

We start with the equation:
\[ xe^{xy} - y = \sin^2 x \]

Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}(xe^{xy}) - \frac{dy}{dx} = \frac{d}{dx}(\sin^2 x) \]

### Step 2: Apply the product rule and chain rule

Using the product rule on \(xe^{xy}\):
\[ \frac{d}{dx}(xe^{xy}) = e^{xy} + x \cdot e^{xy} \cdot \frac{d}{dx}(xy) \]
Now, apply the chain rule to differentiate \(xy\):
\[ \frac{d}{dx}(xy) = y + x \frac{dy}{dx} \]
Thus, we have:
\[ \frac{d}{dx}(xe^{xy}) = e^{xy} + x e^{xy} \left(y + x \frac{dy}{dx}\right) \]

Now substituting this back into our differentiation gives:
\[ e^{xy} + x e^{xy} \left(y + x \frac{dy}{dx}\right) - \frac{dy}{dx} = 2 \sin x \cos x \]

### Step 3: Rearranging the equation

Rearranging gives:
\[ e^{xy} + x e^{xy} y + x^2 e^{xy} \frac{dy}{dx} - \frac{dy}{dx} = 2 \sin x \cos x \]

### Step 4: Collect terms involving \(\frac{dy}{dx}\)

Grouping the terms with \(\frac{dy}{dx}\):
\[ \left(x^2 e^{xy} - 1\right) \frac{dy}{dx} = 2 \sin x \cos x - e^{xy} - x e^{xy} y \]

### Step 5: Solve for \(\frac{dy}{dx}\)

Thus, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = \frac{2 \sin x \cos x - e^{xy} - x e^{xy} y}{x^2 e^{xy} - 1} \]

### Step 6: Evaluate at \(x = 0\)

Now we need to find \(\frac{dy}{dx}\) at \(x = 0\). First, we need to determine \(y\) when \(x = 0\):
Substituting \(x = 0\) into the original equation:
\[ 0 \cdot e^{0 \cdot y} - y = \sin^2(0) \implies -y = 0 \implies y = 0 \]

Now substituting \(x = 0\) and \(y = 0\) into \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{2 \sin(0) \cos(0) - e^{0} - 0 \cdot e^{0} \cdot 0}{0^2 e^{0} - 1} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{0 - 1 - 0}{0 - 1} = \frac{-1}{-1} = 1 \]

### Final Answer

Thus, \(\frac{dy}{dx}\) at \(x = 0\) is:
\[ \boxed{1} \]