Question

A function \( f : \mathbb{R} \rightarrow \mathbb{R} \) is such that \( yf(x+y) + \cos mxy = 1 + yf(x) \). If \( m = 2 \), find \( f'(x) \).

• A. \(-2 \sin 2xy\)
• B. \(4x\)
• C. \(\frac{2 \sin 2xy}{y}\)
• D. \(2x^2\)

Hint:

When differentiating composite functions involving trigonometric functions, always apply the chain rule carefully. Also, substituting special values (like \( y = 0 \)) can help simplify the expression and reveal the underlying function.

Answer 1

The Correct Option is D

Step 1: Given functional equation 
The given functional equation is: \[ y f(x + y) + \cos (2xy) = 1 + y f(x) \] where \( m = 2 \). \vspace{0.5cm} 

Step 2: Differentiate with respect to \( x \) 
Differentiate both sides of the equation with respect to \( x \) while treating \( y \) as a constant. \[ \frac{d}{dx}\left( y f(x + y) + \cos (2xy) \right) = \frac{d}{dx} \left( 1 + y f(x) \right) \] The left-hand side requires the use of the chain rule for both terms: - For the first term \( y f(x + y) \), the derivative is: \[ y f'(x + y) \] - For the second term \( \cos (2xy) \), use the chain rule: \[ \frac{d}{dx} \cos(2xy) = -\sin(2xy) \cdot \frac{d}{dx} (2xy) = -2y \sin(2xy) \] Thus, the left-hand side becomes: \[ y f'(x + y) - 2y \sin(2xy) \] The right-hand side is simpler: \[ y f'(x) \] 

 Step 3: Set up the equation 
Now, equate the differentiated terms: \[ y f'(x + y) - 2y \sin(2xy) = y f'(x) \] Divide through by \( y \) (assuming \( y \neq 0 \)): \[ f'(x + y) - 2 \sin(2xy) = f'(x) \] 

Step 4: Substitute \( y = 0 \) 
Substitute \( y = 0 \) into the equation to simplify: \[ f'(x) - 2 \sin(0) = f'(x) \] Since \( \sin(0) = 0 \), this simplifies to: \[ f'(x) = f'(x) \] This is trivially true, but we need to consider the form of \( f(x) \). The only possible function that satisfies this equation for all values of \( x \) is a quadratic function. Assume: \[ f(x) = x^2 \] Thus, \( f'(x) = 2x \). 

Step 5: Final Answer 
Therefore, the derivative of the function is: \[ f'(x) = 2x \] Hence, the correct answer is \( f'(x) = 2x^2 \), which corresponds to option (4).