If ` xe^(xy)= y+ sin^2 x ` then at ` x=0 , ( dy)/(dx) =. . . .`

Answer

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 of the equation
We start with the equation:
\[ xe^{xy} = y + \sin^2 x \]
We will differentiate both sides with respect to \(x\).

Using the product rule and chain rule on the left-hand side:
\[ \frac{d}{dx}(xe^{xy}) = e^{xy} + x e^{xy} \left( y + x \frac{dy}{dx} \right) \]
For the right-hand side:
\[ \frac{d}{dx}(y + \sin^2 x) = \frac{dy}{dx} + 2\sin x \cos x = \frac{dy}{dx} + \sin(2x) \]

### Step 2: Set the derivatives equal to each other
Now we have:
\[ e^{xy} + x e^{xy} \left( y + x \frac{dy}{dx} \right) = \frac{dy}{dx} + \sin(2x) \]

### Step 3: Substitute \(x = 0\)
Now we will substitute \(x = 0\) into the differentiated equation. First, we need to evaluate \(y\) when \(x = 0\):
\[ 0 \cdot e^{0 \cdot y} = y + \sin^2(0) \implies 0 = y + 0 \implies y = 0 \]

Now substituting \(x = 0\) and \(y = 0\) into the differentiated equation:
\[ e^{0} + 0 \cdot e^{0} \left( 0 + 0 \cdot \frac{dy}{dx} \right) = \frac{dy}{dx} + \sin(0) \]
This simplifies to:
\[ 1 = \frac{dy}{dx} + 0 \]

### Step 4: Solve for \(\frac{dy}{dx}\)
From the equation:
\[ 1 = \frac{dy}{dx} \]
Thus, we have:
\[ \frac{dy}{dx} = 1 \]

### Final Answer
Therefore, at \(x = 0\), \(\frac{dy}{dx} = 1\).

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