Question

If \(f(x) = cosx\,cos2x\,cos4x\,cos8x\,cos16x\), then \(f^'(\frac{π}{4}) =\)

• A. \(\text{cosec}(\frac{π}{4})\)
• B. \(cos(\frac{π}{4})\)
• C. \(tan(\frac{π}{4})\)
• D. \(0\)

Hint:

Simplify the root to tan(x + pi/4) using sin 2x = 2 sin x cos x.

Answer 1

The Correct Option is A

Step 1: Plug in a convenient identity:
Divide numerator and denominator inside the root by $\cos^2 x$-type terms: $\frac{1 + \sin2x}{1 - \sin2x} = \tan^2\left(\frac{\pi}{4} + x\right)$.

Step 2: Result:
So $y = \frac{\pi}{4} + x$ in the neighbourhood of $\frac{\pi}{6}$. Its derivative is 1 (B).

Final Answer:
1. \[ \boxed{1} \]