\(f(x)=\cos x \cdot \cos 2 x \cdot \cos 4 x \cdot \cos 8 x \cdot \cos 16 x\)
\(\because f(x)=\frac{\sin 32 x}{2^{5} \sin x}\)
भाग के नियम से,
\(\frac{d y}{d x}=\frac{v \cdot \frac{d u}{d x}-u \cdot \frac{d v}{d x}}{v^{2}}\)
\(=\frac{32 \sin x \cdot 32 \cos 32 x-32 \cos x \cdot \sin 32 x}{32 \alpha \cdot 32 \sin ^{2} 2 x}\)
\(\therefore f^{\prime}(x)=\frac{32 \sin x \cos 32 x-\cos x \sin 32 x}{32 \sin ^{2} 2 x}\)
अब, \(f^{\prime}\left(\frac{\pi}{4}\right)=\frac{16 \sqrt{2}}{6}=\sqrt{2}\)
