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 find \( y'(0) \) for the equation \( e^{xy^2} + y \cos(x^2) = 5 \), we will follow these steps:

### Step 1: Substitute \( x = 0 \) into the equation
We start by substituting \( x = 0 \) into the equation to find the value of \( y \).

\[ e^{0 \cdot y^2} + y \cos(0^2) = 5 \]

This simplifies to:

\[ e^0 + y \cdot 1 = 5 \]

\[ 1 + y = 5 \]

So, we find:

\[ y = 4 \]

### Step 2: Differentiate the equation implicitly
Next, we differentiate the equation \( e^{xy^2} + y \cos(x^2) = 5 \) with respect to \( x \).

Using the product and chain rules, we differentiate each term:

1. For \( e^{xy^2} \):
\[ \frac{d}{dx}(e^{xy^2}) = e^{xy^2} \left( y^2 + x \cdot 2y \frac{dy}{dx} \right) \]

2. For \( y \cos(x^2) \):
\[ \frac{d}{dx}(y \cos(x^2)) = \frac{dy}{dx} \cos(x^2) + y \cdot \frac{d}{dx}(\cos(x^2)) = \frac{dy}{dx} \cos(x^2) - y \cdot 2x \sin(x^2) \]

Putting it all together, we have:

\[ e^{xy^2} \left( y^2 + 2xy \frac{dy}{dx} \right) + \frac{dy}{dx} \cos(x^2) - 2xy \sin(x^2) = 0 \]

### Step 3: Substitute \( x = 0 \) and \( y = 4 \) into the differentiated equation
Now we substitute \( x = 0 \) and \( y = 4 \) into the differentiated equation to find \( y'(0) \).

Substituting these values:

1. \( e^{0 \cdot 4^2} = e^0 = 1 \)
2. \( y^2 = 4^2 = 16 \)
3. \( \cos(0^2) = \cos(0) = 1 \)
4. \( \sin(0^2) = \sin(0) = 0 \)

The differentiated equation becomes:

\[ 1 \left( 16 + 0 \cdot 2 \cdot 4 \frac{dy}{dx} \right) + \frac{dy}{dx} \cdot 1 - 0 = 0 \]

This simplifies to:

\[ 16 + \frac{dy}{dx} = 0 \]

### Step 4: Solve for \( y'(0) \)
Now, we solve for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = -16 \]

Thus, we find:

\[ y'(0) = -16 \]

### Final Answer
Therefore, the value of \( y'(0) \) is \( -16 \).

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