Question

Find the value of $\frac{d^{20}}{d{x}^{20}}\left(2\cos x\cos 3x\right)$ at $x=\pi$

• A. $2^{10}(2^{10}+1)$
• B. $2^{20}(2^{40}+1)$
• C. $2^{20}(2^{20}+1)$
• D. $2^{40}(2^{20}+1)$

Answer 1

Answer: (C)

$2^{20}(2^{20}+1)$

We have,

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

We know that

$2\cos A\cos B=[\cos (A-B)+\cos (A+B)]$

Therefore,

$=\frac{d^{20}}{d{x}^{20}}[\cos (x-3x)+\cos (x+3x)]$

$=\frac{d^{20}}{d{x}^{20}}[\cos (-2x)+\cos \left(4x\right)]$

$=\frac{d^{20}}{d{x}^{20}}[\cos \left(4x\right)+\cos \left(2x\right)]$

Since, the 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.

Therefore, $5\times 4=20th$ derivative of $\cos x$ is also $\cos x$.

So, apply the chain rule, we get

$\frac{d^{20}}{d{x}^{20}}[\cos \left(4x\right)+\cos \left(2x\right)]=4^{20}\cos 4x+2^{20}\cos 2x$

Since, $x=\pi$

Thus,

$=4^{20}\cos 4\pi +2^{20}\cos 2\pi$

$=4^{20}\times 1+2^{20}\times 1$

$=2^{20}\times 2^{20}+2^{20}$

$=2^{20}(2^{20}+1)$

Hence, this is the answer.