Question

Find the n^th derivative of the following : sin (ax + b)

Answer 1

Let y = sin (ax + b)

Then `"dy"/"dx" = "d"/"dx"[sin(ax + b)]`

= `cos(ax + b)."d"/"dx"(ax + b)`

= cos (ax + b) x (a x 1 + 0)

= `asin[pi/2 + (ax + b)]`

`(d^2y)/(dx^2) = "d"/"dx"[acos(ax + b)]`

= `a"d"/"dx"[cos (ax + b)]`

= `a[- sin(ax + b)]."d"/"dx"(ax + b)`

= a[ – sin (ax + b)] x (a x 1 + 0)
= a² . sin[π + (ax + b)]

= `a^2.sin[(2pi)/2 + (ax + b)]`

`(d^3y)/(dx^3) = "d"/"dx"[-a^2sin(ax + b)]`

= `-a^2"d"/"dx"[sin(ax + b)]`

= `-a^2.cos(ax + b)."d"/"dx"(ax + b)`

= – a². cos(ax + b) x (a x 1 + 0)

= `a^3.sin[(3pi)/2 + (ax + b)]`
In general, the n^th order derivative is given by
`(d^ny)/(dx^n) = a^n.sin[(npi)/2 + (ax + b)]`.