If `int_(0)^(y)e^(-t^(2))dt+int_(0)^(x^(2))sin^(2)tdt=0`, then `(dy)/(dx)` at `x=y=1` is

Options

A. `sin^(-1)`
B. `-esin^(2)1`
C. `-2esin^(2)1`
D. none of these

Answer

To solve the problem, we need to differentiate the given equation with respect to \( x \) and then evaluate \( \frac{dy}{dx} \) at \( x = y = 1 \).

### Step-by-Step Solution

1. **Given Equation**:
\[ \int_{0}^{y} e^{-t^2} dt + \int_{0}^{x^2} \sin^2 t \, dt = 0 \]

2. **Differentiate Both Sides with Respect to \( x \)**:
Using Leibniz's rule for differentiation under the integral sign, we differentiate the left-hand side:
\[ \frac{d}{dx} \left( \int_{0}^{y} e^{-t^2} dt \right) + \frac{d}{dx} \left( \int_{0}^{x^2} \sin^2 t \, dt \right) = 0 \]

3. **Applying Leibniz's Rule**:
For the first integral:
\[ \frac{d}{dx} \left( \int_{0}^{y} e^{-t^2} dt \right) = e^{-y^2} \frac{dy}{dx} \]
For the second integral:
\[ \frac{d}{dx} \left( \int_{0}^{x^2} \sin^2 t \, dt \right) = \sin^2(x^2) \cdot \frac{d}{dx}(x^2) = \sin^2(x^2) \cdot 2x \]

4. **Setting Up the Equation**:
Combining both results, we have:
\[ e^{-y^2} \frac{dy}{dx} + 2x \sin^2(x^2) = 0 \]

5. **Solving for \( \frac{dy}{dx} \)**:
Rearranging the equation gives:
\[ e^{-y^2} \frac{dy}{dx} = -2x \sin^2(x^2) \]
Therefore,
\[ \frac{dy}{dx} = -\frac{2x \sin^2(x^2)}{e^{-y^2}} = -2x \sin^2(x^2) e^{y^2} \]

6. **Evaluating at \( x = 1 \) and \( y = 1 \)**:
Substitute \( x = 1 \) and \( y = 1 \):
\[ \frac{dy}{dx} \bigg|_{x=1, y=1} = -2(1) \sin^2(1^2) e^{1^2} = -2 \sin^2(1) e \]

### Final Answer:
\[ \frac{dy}{dx} \bigg|_{x=1, y=1} = -2 e \sin^2(1) \]