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Question
Find the nth derivative of the following : `(1)/(3x - 5)`
Solution
Let y = `(1)/(3x - 5)`
Then `"dy"/"dx" = "d"/"dx"(3x - 5)`
= `-1(3x - 5)^-2."d"/"dx"(3x - 5)`
= `(-1)/(3x - 5)^2 xx (3 xx 1 - 0)`
= `((-1)^1 .3)/(3x - 5)^2`
`(d^2y)/(dx^2) = "d"/"dx"[((-1)^1 .3)/(3x - 5)^2]`
= `(-1)^1 .3"d"/"dx"(3x - 5)^-2`
= `(-1)^-1 .3.(-2)(3x - 5)^-3."d"/"dx"(3x - 5)`
= `((-1)^2 .3.2)/(3x - 5)^3 xx (3 xx 1 - 0)`
= `((-1)^2. 2!.3^2)/(3x - 5)^3`
`(d^3y)/(dx^3) = "d"/"dx"[((-1)^2. 2!.3^2)/(3x - 5)^3]`
= `(-1)^2 .2!.3^2."d"/"dx"(3x - 5)^-3`
= `(-1)^2 .2!.3^2.(-3)(3x - 5)^-4."d"/"dx"(3x - 5)`
= `((-1)^3 xx 3.2! xx 3^2)/(3x - 5)^4 xx (3 xx 1 - 0)`
= `((-1)^3 xx 3! xx 3^3)/(3x - 5)^4`
In general, the nth order derivative is goven by
`(d^ny)/(dx^n) = ((-1)^n .n!.3^n)/(3x - 5)^(n + 1)`.
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