Question

Find the n^th derivative of the following : y = e^ax . cos (bx + c)

Answer 1

y = e^ax . cos (bx + c)

∴ `"dy"/"dx" = "d"/"dx"[e^(ax).cos[bx + c)]`

= `e^(ax)."d"/"dx"[cos(bx + c)] + cos(bx + c)."d"/"dx"(e^(ax))`

= `e^(ax).[-sin(bx + c)]."d"/"dx"(bx + c) + cos(bx + c).e^(ax)."d"/"dx"(ax)`

= – e^ax sin (bx + c) x (b x 1 + 0) + e^axcos(bx + c) x a x 1

= e^ax [a cos (bx + c) – b sin (bx + c)]

= `e^(ax).sqrt(a^2 + b^2)[a/sqrt(a^2 + b^2)cos(bx + c) - b/sqrt(a^2 + b^2)sin(bx + c)]`

Let `a/sqrt(a^2 + b^2) = cos x and b/sqrt(a^2 + b^2) = sin x`

Then tan ∞ = `b/a`

∴ ∞ = `tan^-1(b/a)`

∴ `"dy"/"dx" = e^(ax).sqrt(a^2 + b^2)[cos∞.cos(bx + c) - sin∞.sin(bx + c)]`

= `e^(ax).(a^2 + b^2)^(1/2).cos(bx + c + x)`

`(d^2y)/(dx^2) = "d"/"dx"[e^(ax).(a^2 + b^2)^(1/2).cos(bx + c + ∞)]`

= `(a^2 + b^2)^(1/2)."d"/"dx"[e^(ax).cos(bx + c + ∞)]`

= `(a^2 + b^2)^(1/2)[e^(ax)."d"/"dx"{cos(bx + c + ∞)} + cos(bx + c + ∞)."d"/"dx"(e^(ax))]`

= `(a^2 + b^2)^(1/2)[e^(ax).{-sin(bx + c + ∞)}."d"/"dx"(bx + c + ∞) + cos(bx +c + ∞).e^(ax)."d"/"dx"(ax)]`

= `(a^2 + b^2)^(1/2)[-e^(ax)sin(bx + c + ∞) xx (b xx 1 + 0 + 0) + cos(bx + c + ∞).e^(ax) xx a xx 1]`

= `e^(ax).(a^2 + b^2)^(1/2)[a cos (bx + c + ∞) - bsin(bx + c + ∞)]`

= `e^(ax).(a^2 + b^2)^(1/2).sqrt(a^2 + b^2)[a/sqrt(a^2 + b^2)cos(bx + c + ∞) = b/sqrt(a^2 + b^2)sin(bx + c + ∞)]`

= `e^(ax).(a^2 + b^2)^(2/2)[cos∞.cos(bx + c + ∞) - sin∞.sin(bx + c + ∞)`

= `e^(ax).(a^2 + b^2)^(2/2).cos(bx + c + ∞ + ∞)`

= `e^(ax).(a^2 + b^2)^(2/2).cos(bx + c + 2∞)`
Similarly.
`(d^3y)/(dx^3) = e^(ax).(a^2 + b^2)^(3/2).cos(bx + c + 3∞)`
In general, the n^th order derivative is given by
`(d^ny)/(dx^n) = e^(ax).(a^2 + b^2)^(n/2).cos(bx + c + n oo)`,

Where ∞ = `tan^-1(b/a)`

∴ `(d^ny)/(dx^n) = e^(ax).(a^2 + b^2)^(n/2).cos[bx + c + ntan^-1(b/a)]`.