Finding the nth derivative of a function using standard forms

Question

$Y = \cos x \cos2x \cos3x$. How to find $Y_n$ ? We can use some standard forms like $Y_n$ for $Y=\cos(ax+b)$ ; $Y_n = a^n \cos[n(\pi/2)+(ax+b)]$ , wherever applicable.

Answer 1

Hint:

Use Werner's formula

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

to find summands of the form $\cos(ax+b)$

Answer 2

$$Y(x)=\cos (x) \cos (2 x) \cos (3 x)=\frac{1}{4} (1+\cos (2 x)+\cos (4 x)+\cos (6 x))$$

$Y_1(x)=\frac{1}{4} (-2 \sin (2 x)-4 \sin (4 x)-6 \sin (6 x))$

$Y_2(x)=\frac{1}{4} (-4 \cos (2 x)-16 \cos (4 x)-36 \cos (6 x))$

$Y_3(x)=\frac{1}{4} (8 \sin (2 x)+64 \sin (4 x)+216 \sin (6 x))$

$Y_4(x)=\frac{1}{4} (16 \cos (2 x)+256 \cos (4 x)+1296 \cos (6 x))$

$\ldots\ldots\ldots\ldots\ldots\ldots$

$Y_{4k+1}=\frac{1}{4} ((-2)^{4k+1} \sin (2 x)+(-4)^{4k+1} \sin (4 x)+(-6)^{4k+1} \sin (6 x))$

$Y_{4k+2}=\frac{1}{4} (-2^{4k+2} \cos (2 x)-4^{4k+2} \cos (4 x)-6^{4k+2} \cos (6 x))$

$Y_{4k+3}=\frac{1}{4} (2^{4k+3} \sin (2 x)+4^{4k+3} \sin (4 x)+6^{4k+3}\sin (6 x))$

$Y_{4k+4}=\frac{1}{4} (2^{4k+4} \cos (2 x)+4^{4k+4} \cos (4 x)+6^{4k+4}\cos (6 x))$

Hope this is useful