Question

Find the n^th derivative of the following : e^ax+b 

Answer 1

Let y = e^ax+b 

Then `"dy"/"dx" = "d"/"dx"(e^(ax + b))`

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

= `e^(ax + b) xx (a xx 1 + 0)`
= ae^ax+b 

`(d^2y)/(dx^3) = "d"/"dx"(ae^(ax + b))`

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

= `ae^(ax + b) xx (a xx 1 + 0)`
= a².e^ax+b

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

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

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

= a²e^ax+b x (a x 1 + 0)
= a³.e^ax+b 
In genaral, the nth order derivative is given by
`(d^ny)/(dx^n)` = aⁿ . e^ax+b.