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प्रश्न
Find the nth derivative of the following : (ax + b)m
उत्तर
Let y = (ax + b)m
Then `"dy"/"dx" = "d"/"dx"(ax + b)^m`
= `m(ax + b)^(m-1)."d"/"dx"(ax + b)`
= m(ax + b)m–1 x (a x 1 + 0)
= am(ax + b)m–1
`(d^2y)/(dx^2) = "d"/"dx"[am(ax + b)^(m-1)]`
= `am"d"/"dx"(ax + b)^(m - 1)`
= `am(m - 1)(ax + b)^(m - 2)."d"/"dx"(ax + b)`
= am(m – 1)(ax + b)m–2 x (a x 1 + 0)
= a2m(m –1) (ax + b)m–2
`(d^2y)/(dx^3) = "d"/"dx"[a^2m(m - 1)"d"/"dx"(ax + b)^(m - 2)]`
= `a^2m(m - 1)"d"/"dx"(ax + b)^(m - 2)`
= `a^m(m - 1)(m - 2)(ax + b)^(m - 3)"d"/"dx"(ax + b)`
= a2m(m – 1)(m – 2)(ax + b)m–3 x (a x 1 + 0)
= a3m(m – 1)(m – 2)(ax + b)m–3
In general, the nth order derivative is given by
`(d^ny)/(dx^n) = a^nm(m - 1)(m - 2)` ...(m – n + 1)(ax + b)m–n
Case (i) : if m > 0, m > n, then
`(d^ny)/(dx^n) = ((a^n.m(m - 1)(m - 2)...(m - n + 1)(m - n)...3.2.1))/((m - n)(m - n - 1)...3.2.1) xx (ax + b)^(m - n)`
∴ `(d^2y)/(dx^n) = ((a^n.m!(ax + b)^(m - n)))/((m - n)!`,If m > 0, m > n.
Case (ii) : if m > 0 and m < n, then its mth order derivative is a constant and every derivatives after mth order are zero.
∴ `(d^ny)/(dx^n)` = 0, if m > 0, m = n.
Case (iii) : If m > 0, m = n, then
`(d^ny)/(dx^n) = a^n . n(n - 1)(n - 2)...(n - n + 1)(ax + b)^(n - n)`
= an.n(n – 1)(n – 2) ... 1.(ax + b)0
∴ `(d^ny)/(dx^n)` = an . n!, if m > 0, m = n.
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