Differentiability
Derivative of Limit
MMTS_Full_Test_03
Grade 12
Question:
If $f(x)=\lim_{n\to\infty}\left(\prod_{i=1}^n\cos\frac{x}{2^i}\right)$ then $f'(x)$ is equal to
$\dfrac{\sin x}{x}$
$\dfrac{x}{\sin x}$
$\dfrac{x\cos x-\sin x}{x^2}$
$\dfrac{\sin x-x\cos x}{\sin^2 x}$
Step-by-Step Solution
Key Concept: Product telescopes to $\sin x / (x\cdot 2^0\cdots)$; use the identity $\prod\cos(x/2^k)=\sin x/x$
$f(x)=\frac{\sin x}{x}$. $f'(x)=\frac{x\cos x-\sin x}{x^2}$.
Correct Answer: 3