$n$-th derivative of $(ax+b)^{-m}$
Solution 1
Let $f(x) = (ax+b)^{-m}\;,$ Then $$f'(x) = -m\cdot (ax+b)^{-(m+1)}\cdot a$$
and $$f^{(2)}(x)=f''(x) = m\cdot (m+1)(ax+b)^{-(m+2)}\cdot a^2$$
similarly $$f^{(3)}(x) = -m\cdot (m+1)\cdot (m+2)(ax+b)^{-(m+3)}\cdot a^3$$
Similarly we can Calculate
$$f^{(n)}(x) = (-1)^n\cdot m\cdot (m+1)\cdot ...(m+n-1)\cdot (ax+b)^{-(m+n)}\cdot a^n$$
So $$\displaystyle f^{(n)}(x) = (-1)^n\cdot \frac{(m+n-1)!}{(m-1)!}\cdot (ax+b)^{-(m+n)}\cdot a^n$$
Solution 2
Assuming $m>0$, let $f_m(x) = \left(ax + b\right)^{-m}$. $$f_m'(x) = -m\left(ax+b\right)^{-m-1} = (-m)\cdot a\cdot f_{m+1}(x)$$ Using this relationship, you can easily apply induction to show that: $$\frac{d^n}{dx^n}f_m(x) = (-1)^na^{n}\frac{(n+m-1)!}{(m-1)!}f_{m+n}(x) = (-1)^na^n\frac{(n+m-1)!}{(m-1)!}\left(ax+b\right)^{-(m+n)}$$
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Srijan Pandey
Updated on October 13, 2022Comments
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Srijan Pandey about 1 year
How to find the $n$-th derivative of $(ax+b)^{-m}$ ?
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abiessu about 8 yearsWhat is the first derivative of the given expression?
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Karl about 8 yearsHint: Differentiate once and see what happens.
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Srijan Pandey about 8 yearsYes I have tried bt made a small mistake that differs the ans... Thanks for the solution
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Dominik about 8 yearsYou forgot to multiply by $a$.
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jameselmore about 8 years@Dominik, thanks