Question

Let \(f(x) = \frac{\sin x}{x}\) for \(x \neq 0\). Then the value of \(f'(\frac{\pi}{2})\) is equal to

• A. \(\frac{-4}{\pi^2}\)
• B. \(\frac{4}{\pi}\)
• C. \(\frac{-2}{\pi}\)
• D. \(\frac{2}{\pi}\)
• E. 0

Hint:

Remember to square the entire denominator. A common error is writing \(\pi/4\) instead of \(\pi^2/4\) in the denominator.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We use the quotient rule to differentiate \(f(x)\).

Step 2: Key Formula or Approach:
Quotient Rule: \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\).

Step 3: Detailed Explanation:
Let \(u = \sin x\) and \(v = x\).
\[ f'(x) = \frac{x \cos x - \sin x(1)}{x^2} = \frac{x \cos x - \sin x}{x^2} \]
Evaluate at \(x = \frac{\pi}{2}\):
\[ f'(\frac{\pi}{2}) = \frac{(\frac{\pi}{2}) \cos(\frac{\pi}{2}) - \sin(\frac{\pi}{2})}{(\frac{\pi}{2})^2} \]
Substitute \(\cos(\frac{\pi}{2}) = 0\) and \(\sin(\frac{\pi}{2}) = 1\):
\[ f'(\frac{\pi}{2}) = \frac{(\frac{\pi}{2})(0) - 1}{\frac{\pi^2}{4}} \]
\[ f'(\frac{\pi}{2}) = \frac{-1}{\frac{\pi^2}{4}} = -\frac{4}{\pi^2} \]

Step 4: Final Answer:
The value is \(-\frac{4}{\pi^2}\).