Question

If \(f\) is a derivable function and \[ 2f(\sin x)+f(\cos x)=x \qquad \forall x\in \mathbb{R}, \] then \[ f'(x)= \]

• A. \[ \sin x+\cos x \]
• B. \[ \sin x-\cos x \]
• C. \[ \sqrt{1-x^2} \]
• D. \[ \frac{1}{\sqrt{1-x^2}} \] \bigskip

Hint:

For equations involving both \(f(\sin x)\) and \(f(\cos x)\), differentiate once and then replace \[ x \mapsto \frac{\pi}{2}-x \] to obtain a second equation. Solving the pair usually gives \(f'\).

Answer 1

The Correct Option is D

Concept: Differentiate the given functional equation, then replace \(x\) by \(\pi/2 - x\) to get a system. Solve for \(f'(\sin x)\) and then substitute \(t=\sin x\).

Step 1: \(2f(\sin x) + f(\cos x) = x\). Differentiate: \(2f'(\sin x)\cos x - f'(\cos x)\sin x = 1\). (1)

Step 2: Replace \(x\) by \(\pi/2 - x\): \(2f'(\cos x)\sin x - f'(\sin x)\cos x = 1\). (2)

Step 3: Let \(A=f'(\sin x), B=f'(\cos x)\). \(2A\cos x - B\sin x = 1\), \(-A\cos x + 2B\sin x = 1\). Multiply first by 2: \(4A\cos x - 2B\sin x = 2\). Add to second: \(3A\cos x = 3 \Rightarrow A = 1/\cos x\). So \(f'(\sin x) = 1/\cos x\).

Step 4: Let \(t=\sin x\), \(\cos x = \sqrt{1-t^2}\). Then \(f'(t) = 1/\sqrt{1-t^2}\). Replace \(t\) by \(x\): \(f'(x) = 1/\sqrt{1-x^2}\).

Step 5: Write the final answer. \(\boxed{\frac{1}{\sqrt{1-x^2}}}\)