Question

If $f(x) = \cos x\cos 2x\cos 4x\cos 8x\cos 16x$, then $f'(\dfrac{\pi }{4})$ is equal to:
A. $\sqrt 2 $
B. $\dfrac{1}{{\sqrt 2 }}$
C. $0$
D. $\dfrac{{\sqrt 3 }}{2}$

Answer 1

Hint: Multiply and divide $\cos x\cos 2x\cos 4x\cos 8x\cos 16x$by $2\sin x$ and then use the formula \[2\sin A\cos A = \sin 2A\]. Multiply and divide the numerator in the next step by 2 and use the same formula \[2\sin A\cos A = \sin 2A\]. After simplifying $f(x)$ use the quotient rule and chain rule for differentiation.

Formula Used:
\[2\sin A\cos A = \sin 2A\]
Quotient rule – If $u$ and $v$ are differentiable with respect to x then $\dfrac{d}{{dx}}\dfrac{u}{v} = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$

Complete step by step Solution:
It is given to us that $f(x) = \cos x\cos 2x\cos 4x\cos 8x\cos 16x$.
Multiply and divide $\cos x\cos 2x\cos 4x\cos 8x\cos 16x$ by $2\sin x$ and use the formula \[2\sin A\cos A = \sin 2A\]
\[\dfrac{{2\sin x\cos x\cos 2x\cos 4x\cos 8x\cos 16x}}{{2\sin x}}\] $ = $ $\dfrac{{\sin 2x\cos 2x\cos 4x\cos 8x\cos 16x}}{{2\sin x}}$
Multiply the numerator and denominator by 2
\[\dfrac{{2\sin 2x\cos 2x\cos 4x\cos 8x\cos 16x}}{{4\sin x}}\]\[\]$ = $ $\dfrac{{\sin 4x\cos 4x\cos 8x\cos 16x}}{{4\sin x}}$
Again, multiply the numerator and denominator by 2
$\dfrac{{2\sin 4x\cos 4x\cos 8x\cos 16x}}{{8\sin x}}$ $ = $ $\dfrac{{\sin 8x\cos 8x\cos 16x}}{{8\sin x}}$
Again, multiply the numerator and denominator by 2
$\dfrac{{2\sin 8x\cos 8x\cos 16x}}{{16\sin x}}$ $ = $ $\dfrac{{\sin 16x\cos 16x}}{{16\sin x}}$
Again, multiply the numerator and denominator by 2
$\dfrac{{2\sin 16x\cos 16x}}{{32\sin x}}$ $ = $ $\dfrac{{\sin 32x}}{{32\sin x}}$
Therefore, $f(x) = \dfrac{{\sin 32x}}{{32\sin x}}$
Using the chain rule and quotient rule,
$f'(x) = \dfrac{{32\sin x\dfrac{d}{{dx}}\sin 32x - \sin 32x\dfrac{d}{{dx}}32\sin x}}{{{{(32\sin x)}^2}}}$
$f'(x) = \dfrac{{1024\sin x\cos 32x - 32\sin 32x\cos x}}{{1024{{\sin }^2}x}}$
Dividing the numerator and denominator by 32,
$f'(x) = \dfrac{{32\sin x\cos 32x - \sin 32x\cos x}}{{32{{\sin }^2}x}}$
$f'(\dfrac{\pi }{4}) = \dfrac{{32\sin (\dfrac{\pi }{4})\cos (8\pi ) - \sin (8\pi )\cos (\dfrac{\pi }{4})}}{{32{{\sin }^2}(\dfrac{\pi }{4})}}$
$f'(\dfrac{\pi }{4}) = \dfrac{{32 \times \dfrac{1}{{\sqrt 2 }} \times 1 - 0 \times \dfrac{1}{{\sqrt 2 }}}}{{32 \times \dfrac{1}{2}}} = \dfrac{{32 \times \dfrac{1}{{\sqrt 2 }}}}{{32 \times \dfrac{1}{2}}} = \dfrac{2}{{\sqrt 2 }}$
Therefore, $f'(\dfrac{\pi }{4}) = \sqrt 2 $

Hence, the correct option is (A).

Note: Chain rule – If $F(x) = f(g(x))$ then $F'(x) = f'(g(x)).g'(x)$. Quotient rule – If $u$ and $v$ are differentiable with respect to x then $\dfrac{d}{{dx}}\dfrac{u}{v} = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$. Instead of considering $32\sin x$ as v, we can also take $\dfrac{1}{{32}}$ as a constant and just consider $\sin x$ as v.