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Algebra of Derivatives

If fx=cos x...

Question

If $f (x) = cos x \cdot cos 2 x \cdot cos 4 x \cdot cos 8 x \cdot cos 16 x$ , then $f^{'} (π / 4)$ equals

A

- \sqrt{2}

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B

0

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C

^{1} / \sqrt{2}

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D

cosec

π /_{4}

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Solution

The correct option is D cosec $π /_{4}$
$f (x) = \frac{2 s i n x c o s x c o s 2 x c o s 4 x c o s 2 x c o s 16 x}{2 s i n x}$
$= \frac{2 s i n 2 x c o s 2 x c o s 4 x c o s 8 x c o s 16 x}{2^{2} s i n x}$
$= \frac{2 s i n 4 x c o s 4 x c o s 8 x c o s 16 x}{2^{3} s i n x}$
$= \frac{2 s i n 8 x c o s 8 x c o s 16 x}{2^{4} s i n x}$
$= \frac{s i n 32 x}{2^{5} s i n x}$
$f^{'} (x) = \frac{1}{2^{5}} (\frac{32 s i n 32 x s i n x - s i n 32 x c o s x}{(s i n^{2} x)})$
$f^{'} (\frac{π}{4}) = \frac{1}{2^{5}} ⎛ ⎜ ⎜ ⎝ \frac{32 \times (1) s i n (\frac{π}{4}) - 0}{s i n^{2} (\frac{π}{4})} ⎞ ⎟ ⎟ ⎠$
$= c o s e c (π / 4)$

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