Question

If \( f(x + ay) + g(x - ay) = 0 \), then \( \frac{dy}{dx} = \):

• A. \( \frac{f'(x - ay) + g'(x + ay)}{g'(x + ay) - f'(x - ay)} \)
• B. \( \frac{f'(x + ay) + g'(x - ay)}{g'(x - ay) - f'(x + ay)} \)
• C. \( \frac{f'(x + ay) + g'(x - ay)}{f'(x + ay) + g'(x - ay)} \)
• D. \( \frac{f'(x + ay) + g'(x - ay)}{f'(x + ay) g'(x - ay)} \)

Hint:

Use implicit differentiation and the chain rule to differentiate the equation and solve for \( \frac{dy}{dx} \).

Answer 1

The Correct Option is B

We are given the equation \( f(x + ay) + g(x - ay) = 0 \). Differentiating both sides with respect to \( x \), we apply the chain rule: \[ \frac{d}{dx}\left(f(x + ay)\right) + \frac{d}{dx}\left(g(x - ay)\right) = 0 \] This gives: \[ f'(x + ay) \cdot \frac{d}{dx}(x + ay) + g'(x - ay) \cdot \frac{d}{dx}(x - ay) = 0 \] \[ f'(x + ay) \cdot (1 + a \frac{dy}{dx}) + g'(x - ay) \cdot (1 - a \frac{dy}{dx}) = 0 \] Now, solve for \( \frac{dy}{dx} \): \[ f'(x + ay) + a f'(x + ay) \frac{dy}{dx} + g'(x - ay) - a g'(x - ay) \frac{dy}{dx} = 0 \] Rearrange the terms to isolate \( \frac{dy}{dx} \): \[ a(f'(x + ay) + g'(x - ay)) \frac{dy}{dx} = -(f'(x + ay) + g'(x - ay)) \] Thus: \[ \frac{dy}{dx} = \frac{f'(x + ay) + g'(x - ay)}{g'(x - ay) - f'(x + ay)} \] Thus, the correct answer is option (2).