Question

If \(f(x) = cosxcos2xcos4xcos8xcos16x\), then \(f^'(\frac{π}{4})\) is equal to

• A. \(1\)
• B. \(0\)
• C. \(\sqrt{2}\)
• D. \(\frac{1}{\sqrt{2}}\)

Hint:

Use \(\prod\cos(2^kx)=\dfrac{\sin32x}{32\sin x}\) before differentiating.

Answer 1

The Correct Option is C

Step 1: Product rule idea
At $x=\pi/4$ only $\cos2x$ vanishes, so only the term where $\cos2x$ is differentiated survives.

Step 2: Evaluate
That term is $\cos x\cdot(-2\sin2x)\cdot\cos4x\cos8x\cos16x$. At $\pi/4$: $\frac{1}{\sqrt2}\cdot(-2)(1)\cdot(-1)(1)(1)=\sqrt2$. ($\cos\pi=-1$, $\cos2\pi=1$, $\cos4\pi=1$.) Option (C).

Final Answer:
The derivative is $\sqrt2$, option (C). \[ \boxed{\sqrt2} \]