Question

$\frac{d^{10}}{d{x}^{10}}(\cos 3x.\cos 2x)$ is equal to

• A. $\frac{1}{2}[5^{{}^{10}}\cos 5x+\cos x]$
• B. $\frac{1}{2}[-5^{{}^{10}}\cos 5x-\cos x]$
• C. $\frac{1}{2}[5^{{}^{10}}\cos 5x-\cos x]$
• D. $\frac{1}{2}[-5^{{}^{10}}\cos 5x+\cos x]$

Answer 1

The correct option is B $\frac{1}{2}[-5^{{}^{10}}\cos 5x-\cos x]$

Let $y=\cos 3x\cos 2x$

$2y=2\cos 3x\cos 2x$

$=\cos (3x+2x)+\cos (3x-2x)$

$2y=\cos 5x+\cos x$

differentiate on both sides w.r.t x

$2\frac{dy}{dx}=-5\sin 5x-\sin x$

$2\frac{d^{2}y}{d{x}^2}=-5^2\cos 5x-\cos x$

$2\frac{d^{3}y}{d^{3}x}=5^3\sin 5x+\sin x$

$\therefore 2\frac{d^{4n+2}y}{d{x}^{4n+2}}=-5^{4n+2}\cos 5x-\cos x$

$2\frac{d^{4n}y}{d{x}^{4n}}=5^{4n}\cos 5x+\cos x$

$2\frac{d^{4n+1}y}{d{x}^{4n+1}}=-5^{4n+1}\sin 5x-\sin x$

$2\frac{d^{4n+3}y}{d{x}^{4n+3}}=5^{4n+3}\sin 5x+\sin x$

$10=4\left(2\right)+2$

$\therefore \frac{d^{10}\left(y\right)}{d^{10}x}=\frac{1}{2}[-5^{10}\cos 5x-\cos x]$