If `y=y(x)` and it follows the relation `e^(xy^2)+ycos(x^2)=5` then `y'(0)` is equal to

Options

A. 4
B. `-16`
C. `-4`
D. `16`

Answer

To solve the problem, we need to differentiate the given equation and find the value of \( y'(0) \).

### Step-by-Step Solution:

1. **Given Equation**:
\[ e^{xy^2} + y \cos(x^2) = 5 \]

2. **Differentiate Both Sides with Respect to \( x \)**:
Using implicit differentiation, we differentiate the left side:
\[ \frac{d}{dx}(e^{xy^2}) + \frac{d}{dx}(y \cos(x^2)) = 0 \]

3. **Differentiate \( e^{xy^2} \)**:
Using the product rule and chain rule:
\[ \frac{d}{dx}(e^{xy^2}) = e^{xy^2} \cdot \frac{d}{dx}(xy^2) = e^{xy^2} \left( y^2 + x \cdot 2y \frac{dy}{dx} \right) \]

4. **Differentiate \( y \cos(x^2) \)**:
Again using the product rule:
\[ \frac{d}{dx}(y \cos(x^2)) = \frac{dy}{dx} \cos(x^2) + y \cdot \frac{d}{dx}(\cos(x^2)) = \frac{dy}{dx} \cos(x^2) - y \sin(x^2) \cdot 2x \]

5. **Combine the Derivatives**:
Combining both parts gives:
\[ e^{xy^2} \left( y^2 + 2xy \frac{dy}{dx} \right) + \frac{dy}{dx} \cos(x^2) - 2xy \sin(x^2) = 0 \]

6. **Factor Out \( \frac{dy}{dx} \)**:
Rearranging the equation to isolate \( \frac{dy}{dx} \):
\[ \left( 2xy e^{xy^2} + \cos(x^2) \right) \frac{dy}{dx} = 2xy \sin(x^2) - y^2 e^{xy^2} \]
Thus,
\[ \frac{dy}{dx} = \frac{2xy \sin(x^2) - y^2 e^{xy^2}}{2xy e^{xy^2} + \cos(x^2)} \]

7. **Evaluate at \( x = 0 \)**:
Now we need to find \( y'(0) \). First, we need to find \( y(0) \) by substituting \( x = 0 \) into the original equation:
\[ e^{0 \cdot y(0)^2} + y(0) \cos(0^2) = 5 \]
This simplifies to:
\[ 1 + y(0) = 5 \implies y(0) = 4 \]

8. **Substitute \( x = 0 \) and \( y(0) = 4 \) into the Derivative**:
Now substitute \( x = 0 \) and \( y(0) = 4 \) into the derivative expression:
\[ \frac{dy}{dx} \bigg|_{x=0} = \frac{2 \cdot 0 \cdot 4 \cdot \sin(0) - 4^2 e^{0}}{2 \cdot 0 \cdot 4 \cdot e^{0} + \cos(0)} = \frac{0 - 16}{0 + 1} = -16 \]

9. **Final Result**:
Therefore, the value of \( y'(0) \) is:
\[ y'(0) = -16 \]