Question

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

Answer 1

Let y = log (ax + b)

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

= `(1)/(ax + b)."d"/"dx"(ax + b)`

= `(1)/(ax + b) xx (a xx 1 + 0)`

= `a/"ax + b"`

`(d^2y)/(dx^2) = "d"/"dx"(a/(ax + b))`

= `a"d"/"dx"(ax + b)^-1`

= `a(-1)(ax + b)^-2."d"/"dx"(ax + b)`

= `((-1)a)/((ax + b)^2) xx (a xx 1 + 0)`

= `((-1)a^2)/((ax + b)^2)`

`(d^3y)/(dx^3) = "d"/"dx"[((-1)^1a^2)/(ax + b)^2]`

= `(-1)^1a^2."d"/"dx"(ax + b)^-2`

= `(-1)^1a^2.(-2)(ax + b)^-3."d"/"dx"(ax + b)`

= `((-1)^2. 1.2.a^2)/(ax + b)^3 xx (a xx 1 + 0)`

= `((-1)^2 .2! a^3)/(ax + b)^3`
In general, the n^th order derivative is given by

`(d^ny)/(dx^n) = ((-1)^(n - 1).(n - 1)!a^n)/(ax + b)^n`