Question

Find nth derivative of sinx. Cosx2

Answer 1

let y = (Cos x)^2 * sinx = (1/2)* ( 1+cos 2x) * sinx

=> y= sinx / 2 +sinx * cos(2x)

=> = sinx / 2 + sin x * sin (π/2 -2x)

=> y= (sinx )/2 +(1/2)*[ sin ( π/2-x ) cos (3x- π/2)

y = (1/2)*(y1 +y2 +y3 )

With y1= sinx

; y2 = sin ( π/2 -x) and y3 = cos ( 3x -π/2 )

We have y1 ‘ = cosx = sin ( π/2 +x)

y1″ = cos ( π/2 +x) = sin ( 2π/2 +x)

And y1 (n) = sin ( n π/2 +x)

y2 =sin ( π/2 - x) = cos x

=> y2 ‘ = -sin ( π/2 +x) = cos (π/2 +x)

=> y2″ = -sin ( π/2 +x) = cos (2 π/2 + x)

And y2(n) = cos ( nπ/2 + x)

y3 = cos (3x -π/2 ) = cos ( π/2 -3x) = sin 3x

=> y3' = 3 cos 3x = 3 sin ( π/2 +3x)

=> y3″ = 9 cos ( π/2 + 3x) = 3^2 . sin ( 2 π/2 +3x)

=> y3(n) = 3^n sin ( n π/2 +3x)

Where y(n) = (1/2) [ y1(n) + y2(n) +y3(n) ] nth derivative is =

=> y(n) = d^n y / dx^n = (1/2) [ sin ( n π/2 +x) + cos ( n π/2 +x) + (3^n )sin ( n π/2 +3x) ]