Question

If \[ 3f(\cos x)+2f(\sin x)=5x, \] then \[ f'(\cos x)+f'(\sin x)= \]

• A. \(-5(\sin x+\cos x)\)
• B. \(-5\sin x\cos x\)
• C. \(-\frac{5}{\sin x}-\frac{5}{\cos x}\)
• D. \(\frac{5}{\sin x}+\frac{5}{\cos x}\)

Hint:

When a function contains both \(f(\cos x)\) and \(f(\sin x)\), replace \(x\) by \(\frac{\pi}{2}-x\) to get another useful equation involving \(f(\sin x)\) and \(f(\cos x)\).

Answer 1

The Correct Option is C

Step 1: Write the given equation.
Given, \[ 3f(\cos x)+2f(\sin x)=5x \] Differentiate both sides with respect to \(x\).
\[ 3f'(\cos x)(-\sin x)+2f'(\sin x)(\cos x)=5 \] So, \[ -3\sin x f'(\cos x)+2\cos x f'(\sin x)=5 \]

Step 2: Replace \(x\) by \(\frac{\pi}{2}-x\).
Since, \[ \cos\left(\frac{\pi}{2}-x\right)=\sin x \] and \[ \sin\left(\frac{\pi}{2}-x\right)=\cos x, \] we get, \[ 3f(\sin x)+2f(\cos x)=5\left(\frac{\pi}{2}-x\right) \] Differentiate both sides with respect to \(x\).
\[ 3f'(\sin x)\cos x+2f'(\cos x)(-\sin x)=-5 \] So, \[ -2\sin x f'(\cos x)+3\cos x f'(\sin x)=-5 \]

Step 3: Solve the two equations.
Let \[ A=f'(\cos x) \] and \[ B=f'(\sin x) \] Then, \[ -3\sin x A+2\cos x B=5 \] and \[ -2\sin x A+3\cos x B=-5 \] Solving these two equations, we get \[ A=-\frac{5}{\sin x} \] and \[ B=-\frac{5}{\cos x} \]

Step 4: Find the required value.
Therefore, \[ f'(\cos x)+f'(\sin x) = -\frac{5}{\sin x}-\frac{5}{\cos x} \]

Step 5: Final conclusion.
Hence, \[ \boxed{-\frac{5}{\sin x}-\frac{5}{\cos x}} \]