Question

Find the n^th derivative of the following : cos (3 – 2x)

Answer 1

Let y = cos (3 – 2x)

Differentiating both sides w.r.t. x, we get,

`dy/dx = d/dx [cos (3  –  2x)]`

`dy/dx = – sin(3  –  2x). d/dx (3  –  2x)`

`dy/dx = – sin(3  –  2x) × (0  –  2 × 1)`

`dy/dx = (–  2) (– sin(3  –  2x))`

`dy/dx = (– 2) cos [π/2 + (3  –  2x)]`

Again differentiating both sides w.r.t. x, we get,

`(d^2y)/(dx^2) = d/dx [(–  2) (– sin(3  –  2x)]`

`(d^2y)/(dx^2) = (–  2) d/dx  [(– sin(3  –  2x)]`

`(d^2y)/(dx^2) = (–  2) [– cos (3  –  2x)] d/dx (3  –  2x)`

`(d^2y)/(dx^2) = (–  2) [– cos (3  –  2x)] × (0  –  2 × 1)`

`(d^2y)/(dx^2) = (–  2) [– cos (3  –  2x)] × (–  2)`

`(d^2y)/(dx^2) = (2)^2 [cos ((2π)/2) + (3  –  2x)]`

Again differentiating both sides w.r.t. x, we get,

`(d^3y)/(dx^3) = d/dx [(2)^2 (– cos (3  –  2x))]`

`(d^3y)/(dx^3) = (2)^2 d/dx [– cos (3  –  2x)]`

`(d^3y)/(dx^3) = (2)^2 [sin (3  –  2x)] d/dx (3  –  2x)`

`(d^3y)/(dx^3) = (2)^2 [sin (3  –  2x)] × (0  –  2 × 1)`

`(d^3y)/(dx^3) = (2)^2 [sin (3  –  2x)] (–  2)`

`(d^3y)/(dx^3) = (– 2)^3 [cos ((3π)/2) + (3  –  2x)]`

In general, the n^th order derivative is given by

`(d^ny)/(dx^n) = (-2)^n cos[(nπ)/2 + (3 - 2x)]`.