# $n$-th derivative of $(ax+b)^{-m}$

3,083

## 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, 2022

• How to find the $n$-th derivative of $(ax+b)^{-m}$ ?

• • • • You forgot to multiply by $a$.
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