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Question
Find the nth derivative of the following : log (2x + 3)
Solution
Let y = log (2x + 3)
Then `"dy"/"dx" = "d"/"dx"[log(2x + 3)]`
= `(1)/(2x + 3)."d"/"dx"(2x + 3)`
= `(1)/(2x + 3) xx (a xx 1 + 0)`
= `a/"2x + 3"`
`(d^2y)/(dx^2) = "d"/"dx"(a/(2x + 3))`
= `a"d"/"dx"(2x + 3)^-1`
= `a(-1)(2x + 3)^-2."d"/"dx"(2x + 3)`
= `((-1)a)/((2x + 3)^2) xx (a xx 1 + 0)`
= `((-1)a)/((2x + 3)^2)`
`(d^3y)/(dx^3) = "d"/"dx"[((-1)^1a^2)/(2x + 3)^2]`
= `(-1)^1a^2."d"/"dx"(2x + 3)^-2`
= `(-1)^1a^2.(-2)(2x + 3)^-3."d"/"dx"(2x + 3)`
= `((-1)^2. 1.2.a^2)/(2x + 3)^3 xx (a xx 1 + 0)`
= `((-1)_^2.2! a^3)/(2x + 3)^3`
In general, the nth order derivative is given by
`(d^ny)/(dx^2) = ((-1)^(n - 1).(n - 1)!2^n)/(2x + 3)^n`.
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