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Question
Find the nth derivative of the following:
`(1)/x`
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Solution
Let y = `(1)/x`
Then `"dy"/"dx" = "d"/"dx"(1/x)`
= `-(1)/x^2`
= `((-1)^1 1!)/x^2`
`(d^2y)/(dx^2) = "d"/"dx"(-1/x^2)`
= `1"d"/"dx"(x^-2)`
= ( – 1)(– 2)x–3
= `((-1)^2. 1.2)/x^3`
= `((-1)^2 2!)/x^3`
`(d^3y)/(dx^3) = "d"/"dx"[((-1)^2. 2!)/x^3]`
= `(-1)^2. 2!"d"/"dx"(x^-3)`
= ( –1)2. 2!.( – 3)x–4
= `((-1)^3 xx 3.2!)/x^4`
= `((-1)^3. 3!)/x^4`
In general, the nth order derivative is given by
`(d^ny)/(dx^n) = ((-1)^n. n!)/(x^(n + 1)`.
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