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प्रश्न
Find the nth derivative of the following : log (ax + b)
उत्तर
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 nth order derivative is given by
`(d^ny)/(dx^2) = ((-1)^(n - 1).(n - 1)!a^n)/(ax + b)^n`
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