If f ( x ) = lim n ( i = 1 n cos ( x 2 i ) ) then f ' ( x ) is equal to:

Question

If $f (x) = \lim_{n \to \infty} (\prod_{i = 1}^{n} \cos (\frac{x}{2^{i}}))$ then $f' (x)$ is equal to:

Smart Achievers · Accepted Answer

(3)

Using the identity

$\sin x = 2^{n} \sin (\frac{x}{2^{n}}) \prod_{i = 1}^{n} \cos (\frac{x}{2^{i}})$

we get

$\prod_{i = 1}^{n} \cos (\frac{x}{2^{i}}) = \frac{\sin x}{2^{n} \sin (\frac{x}{2^{n}})}$

As $n \to \infty$ ,

$2^{n} \sin (\frac{x}{2^{n}}) \to x$

Hence,

$f (x) = \frac{\sin x}{x}$

Differentiating,

$f' (x) = \frac{x \cos x - \sin x}{x^{2}}$

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