Question

$\frac{d^{20}\left(2\cos x\cos 3x\right)}{d{x}^{20}}=$

• A. $2^{20}(\cos 2x-2^{20}\cos 4x)$
• B. $-2^{20}(\cos 2x+2^{20}\cos 4x)$
• C. $2^{20}(\sin 2x+2^{20}\sin 4x)$
• D. $2^{20}(\sin 2x-2^{20}\sin 4x)$

Answer 1

The correct option is B $-2^{20}(\cos 2x+2^{20}\cos 4x)$

$Firstletussolvetheinnerfunctionindividually=\frac{d}{dx}\left(2\cos x\cos 3x\right)$
Using this identity,
$2cos\left(x\right)cos\left(3x\right)=cos(-2x)+cos\left(4x\right)=cos\left(2x\right)+cos\left(4x\right).$
So, we need to compute the 20th derivative of cos(2x) + cos(4x).
Since successive derivatives of cos x cycle in 4:
-sin x, -cos x, sin x, cos x, ...
Since the 4th derivative of cos x is cos x itself,
the 20 (= 5 · 4) th derivative of cos x is also cos x.
Keeping the Chain Rule in mind, the answer to your question is
$-2²⁰cos\left(2x\right)-4²⁰cos\left(4x\right).$
$⟹-2^{20}(\cos 2x+2^{20}\cos 4x)$