y $ =cos\;x\;cos\;2x\;cos\;3x \\ \; \\ \therefore y\; = \; \dfrac{1}{2} [ 2cos\;x\;cos\;2x\;cos\;3x] \\ \; \\ \; \\ =\dfrac{1}{2} \Big\{ [ cos(x+2x) \; + \; cos(x-2x) ] \; cos\;3x \Big\} \\ \; \\ \; \\ =\dfrac{1}{2} \Big\{ [ cos\;3x \; + \; cos\;x ] \; cos\;3x \Big\} \; \; \ldots\{ \because cos(-x)=cos\;x\} \\ \; \\ \; \\ =\dfrac{1}{2} \Big\{ cos^23x \; + \; cos\;xcos\;3x \; \Big\} \\ \; \\ \; \\ =\dfrac{1}{4} \Big\{ 2cos^23x \; + \; 2cos\;xcos\;3x \; \Big\} \\ \; \\ \; \\ =\dfrac{1}{4} \Big\{ 1+cos6x \; + \; cos\;4x + cos2x \; \Big\} \\ \; \\ \ldots \{\because 2cosAcosB \; = \; cos(A+B)+cos(A-B)\} \\ \; \\ \; \\ \; \\ Now, \; diff \; wrt \; x, \\ \; \\ \; \\ \therefore y_1 \; = \; =\dfrac{1}{4} \Big[ 0-2sin2x -4sin\;4x -6sin6x \; \Big] \\ \; \\ \; \\ =\dfrac{1}{4} \Big[ -2sin2x -4sin\;4x -6sin6x \; \Big] \; \; \ldots (i) \\ \; \\ \; \\ \therefore y_1 \; = \; \dfrac{1}{4} \Bigg[ -2cos(1\cdot \frac{\pi}{2}-2x) -4cos(1\cdot \frac{\pi}{2}-4x) -6cos(1\cdot \frac{\pi}{2}-6x) \Bigg] $

• This is just an adjustment to get a proper format of $y_n $ equation

$ \therefore cos(\frac{\pi}{2}-\theta) \;=\; -sin\theta$,

we have to multiply the equation by (-1) to keep the value of $ y_n $ same as that before this adjustment.

Diff again w.r.t. x, equation (i),

$ \therefore y_2 \;=\; =\dfrac{1}{4} \Big[ -4cos2x -16cos\;4x -36cos6x \; \Big] \\ \; \\ \; \\ =\dfrac{1}{4} \Big[ -2^2cos2x -4^2cos\;4x -6^2cos6x \; \Big] \\ \; \\ \; \\ \; \\ y_3 \; = \; =\dfrac{1}{4} \Big[ +8sin2x +64sin\;4x +216sin6x \; \Big] \\ \; \\ \; \\ \therefore y_3 \;= \; \dfrac{(-1)^3}{4} \Bigg[ 2^3 \; cos(3 \cdot \frac{\pi}{2}-2x) + 4^3 \; cos(3 \cdot \frac{\pi}{2}-4x) + 6^3 \; cos(3 \cdot \frac{\pi}{2}-6x) \Bigg] \\ \; \\ \; \\ \; \\ y_4 \; = \; =\dfrac{1}{4} \Big[ 16cos2x +4^4sin\;4x +6^4sin6x \; \Big] \\ \; \\ \; \\ = \; \dfrac{(-1)^4}{4} \Bigg[ 2^4 \; cos(4 \cdot \frac{\pi}{2}-2x) + 4^4 \; cos(4 \cdot \frac{\pi}{2}-4x) + 6^4 \; cos(4 \cdot \frac{\pi}{2}-6x) \Bigg] $

Diff w.r.t x ‘n’ times, we get

$$ y_n \; = \; \dfrac{(-1)^n}{4} \Bigg[ 2^n \; cos(n \cdot \frac{\pi}{2}-2x) + 4^n \; cos(n \cdot \frac{\pi}{2}-4x) + 6^n \; cos(n \cdot \frac{\pi}{2}-6x) \Bigg] $$