If `y = a sin x + b cos x," then "y^(2) + ((dy)/(dx))^(2)` =

Options

A. `a^(2)-b^(2)`
B. `a^(2) +b^(2)`
C. `2a^(2)`
D. `2b^(2)`

Answer

To solve the problem, we need to find the expression for \( y^2 + \left( \frac{dy}{dx} \right)^2 \) given that \( y = a \sin x + b \cos x \).

### Step-by-Step Solution:

1. **Write the expression for \( y \)**:
\[ y = a \sin x + b \cos x \]

2. **Calculate \( y^2 \)**:
\[ y^2 = (a \sin x + b \cos x)^2 \]
Expanding this using the formula \( (A + B)^2 = A^2 + B^2 + 2AB \):
\[ y^2 = a^2 \sin^2 x + b^2 \cos^2 x + 2ab \sin x \cos x \]

3. **Calculate \( \frac{dy}{dx} \)**:
\[ \frac{dy}{dx} = \frac{d}{dx}(a \sin x + b \cos x) = a \cos x - b \sin x \]

4. **Calculate \( \left( \frac{dy}{dx} \right)^2 \)**:
\[ \left( \frac{dy}{dx} \right)^2 = (a \cos x - b \sin x)^2 \]
Expanding this:
\[ \left( \frac{dy}{dx} \right)^2 = a^2 \cos^2 x + b^2 \sin^2 x - 2ab \sin x \cos x \]

5. **Add \( y^2 \) and \( \left( \frac{dy}{dx} \right)^2 \)**:
\[ y^2 + \left( \frac{dy}{dx} \right)^2 = \left( a^2 \sin^2 x + b^2 \cos^2 x + 2ab \sin x \cos x \right) + \left( a^2 \cos^2 x + b^2 \sin^2 x - 2ab \sin x \cos x \right) \]

6. **Combine like terms**:
\[ = a^2 \sin^2 x + b^2 \cos^2 x + 2ab \sin x \cos x + a^2 \cos^2 x + b^2 \sin^2 x - 2ab \sin x \cos x \]
The \( 2ab \sin x \cos x \) and \( -2ab \sin x \cos x \) cancel each other out:
\[ = a^2 \sin^2 x + b^2 \sin^2 x + a^2 \cos^2 x + b^2 \cos^2 x \]
\[ = (a^2 + b^2)(\sin^2 x + \cos^2 x) \]

7. **Use the Pythagorean identity**:
Since \( \sin^2 x + \cos^2 x = 1 \):
\[ y^2 + \left( \frac{dy}{dx} \right)^2 = a^2 + b^2 \]

### Final Answer:
\[ y^2 + \left( \frac{dy}{dx} \right)^2 = a^2 + b^2 \]