Question

find nth derivative of cos²x

Answer 1

y1 = d/dx (cos2x)  

Now,

cos2x = (1+cos2x)/2

So, we can just substitute that :

y1 = d/dx((1+cos2x)/2)

y1 = 1/2 d/dx(1)+1/2 d/dx (cos2x)

Let’s make another substitution, let u=2x.

y1 = 0+1/2 d/du (cosu) d/dx(2x)

y1 = −sin2x

We know that −sinx = sin(−x) and sinx = cos(π/2−x).

y1 = sin(−2x)

y1 = cos(π/2+2x)

Then, let’s go for the second derivative :

y2 = d/dx cos(π/2+2x)

Let v = π/2+2x

y2 = d/dv cosv d/dx(π/2+2x)

y2 = d/dv cosv [d/dx(π/2)+d/dx 2x]

y2 = −sin(π/2+2x)[0+2]

y2 = −2sin(π/2+2x)

y2 = 2sin(−[π2+2x])

Again, converting the function into cosines :

y2 = 2cos(π2+π2+2x)

y2 = 2cos(2*π/2+2x)

Now, we have a pattern,

yn = n cos (nπ/2+2x)

That is the n-th derivative.