If `y=y(x)` and it follows the relation `4x e^(x y)=y+5sin^2x ,` then `y^(prime)(0)` is equal to______

Explanation

To find \( y'(0) \) for the relation given by

\[ 4x e^{xy} = y + 5 \sin^2 x, \]

we will differentiate both sides with respect to \( x \) and then evaluate at \( x = 0 \).

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

Using the product rule and the chain rule, we differentiate the left-hand side:

\[ \frac{d}{dx}(4x e^{xy}) = 4 \cdot e^{xy} + 4x \cdot e^{xy} \cdot (y + xy'). \]

For the right-hand side, we differentiate:

\[ \frac{d}{dx}(y + 5 \sin^2 x) = y' + 5 \cdot 2 \sin x \cos x = y' + 5 \sin(2x). \]

### Step 2: Set the derivatives equal

Now we set the derivatives equal to each other:

\[ 4 e^{xy} + 4x e^{xy} (y + xy') = y' + 5 \sin(2x). \]

### Step 3: Evaluate at \( x = 0 \)

Now we will evaluate both sides at \( x = 0 \):

- For the left-hand side:
- \( e^{xy} \) at \( x = 0 \) becomes \( e^{0} = 1 \).
- The term \( 4x e^{xy} (y + xy') \) becomes \( 0 \) since \( x = 0 \).

Thus, the left-hand side becomes:

\[ 4 \cdot 1 + 0 = 4. \]

- For the right-hand side:
- \( \sin(2x) \) at \( x = 0 \) becomes \( 0 \).

Thus, the right-hand side becomes:

\[ y' + 0 = y'. \]

### Step 4: Set the equations equal

Now we have:

\[ 4 = y'. \]

### Step 5: Conclusion

Therefore, we find that:

\[ y'(0) = 4. \]

### Final Answer

Thus, \( y'(0) \) is equal to \( \boxed{4} \).

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