Question

Find the nth derivative of (a) y=cos8x cos4x cos2x

Answer 1

Answer:

Step-by-step explanation:

We have,

$\rm{y=cos(8x)\,cos(4x)\,cos(2x)}$

$\rm{\implies\,y=\dfrac{1}{2}\cdot2\,cos(8x)\,cos(4x)\cdot\,cos(2x)}$

$\rm{\implies\,y=\dfrac{1}{2}\left\{cos\left(\dfrac{8x+4x}{2}\right)+cos\left(\dfrac{8x-4x}{2}\right)\right\}\cdot\,cos(2x)}$

$\rm{\implies\,y=\dfrac{1}{2}\left\{cos(6x)+cos(2x)\right\}\cdot\,cos(2x)}$

$\rm{\implies\,y=\dfrac{1}{2}\,cos(6x)\,cos(2x)+\dfrac{1}{2}\,cos(2x)\,cos(2x)}$

$\rm{\implies\,y=\dfrac{1}{4}\cdot2\,cos(6x)\,cos(2x)+\dfrac{1}{2}\,cos^{2}(2x)}$

$\rm{\implies\,y=\dfrac{1}{4}\left\{cos\left(\dfrac{6x+2x}{2}\right)+cos\left(\dfrac{6x-2x}{2}\right)\right\}+\dfrac{1}{2}\left\{1+cos(4x)\right\}}$

$\rm{\implies\,y=\dfrac{1}{4}\left\{cos(4x)+cos(2x)\right\}+\dfrac{1}{2}\left\{1+cos(4x)\right\}}$

$\rm{\implies\,y=\dfrac{1}{4}\,cos(4x)+\dfrac{1}{4}\,cos(2x)+\dfrac{1}{2}+\dfrac{1}{2}\,cos(4x)}$

$\rm{\implies\,y=\left(\dfrac{1}{4}+\dfrac{1}{2}\right)cos(4x)+\dfrac{1}{4}\,cos(2x)+\dfrac{1}{2}}$

$\rm{\implies\,y=\dfrac{3}{4}\,cos(4x)+\dfrac{1}{4}\,cos(2x)+\dfrac{1}{2}}$

We know,

$\boxed{\tt{\dfrac{d^{n}}{dx^{n}}\left\{cos(a\,x+b)\right\}={a}^{n}\cdot\,cos\left(a\,x+b+\dfrac{n\pi}{2}\right)}}$

Now,

$\rm{\implies\,y_{n}=\dfrac{3}{4}\cdot\,{4}^{n}\,cos\left(4x+\dfrac{n\pi}{2}\right)+\dfrac{1}{4}\cdot{2}^{n}\,cos\left(2x+\dfrac{n\pi}{2}\right)+0}$

$\rm{\implies\,y_{n}=3\cdot\,{4}^{n-1}\,cos\left(4x+\dfrac{n\pi}{2}\right)+{2}^{n-2}\,cos\left(2x+\dfrac{n\pi}{2}\right)}$