$n$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
$$f^{(n)}(x) = (1)^n\cdot m\cdot (m+1)\cdot ...(m+n1)\cdot (ax+b)^{(m+n)}\cdot a^n$$
So $$\displaystyle f^{(n)}(x) = (1)^n\cdot \frac{(m+n1)!}{(m1)!}\cdot (ax+b)^{(m+n)}\cdot a^n$$
Solution 2
Assuming $m>0$, let $f_m(x) = \left(ax + b\right)^{m}$. $$f_m'(x) = m\left(ax+b\right)^{m1} = (m)\cdot a\cdot f_{m+1}(x)$$ Using this relationship, you can easily apply induction to show that: $$\frac{d^n}{dx^n}f_m(x) = (1)^na^{n}\frac{(n+m1)!}{(m1)!}f_{m+n}(x) = (1)^na^n\frac{(n+m1)!}{(m1)!}\left(ax+b\right)^{(m+n)}$$
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Srijan Pandey
Updated on October 13, 2022Comments

Srijan Pandey about 1 year
How to find the $n$th derivative of $(ax+b)^{m}$ ?

abiessu about 8 yearsWhat is the first derivative of the given expression?

Karl about 8 yearsHint: Differentiate once and see what happens.

Srijan Pandey about 8 yearsYes I have tried bt made a small mistake that differs the ans... Thanks for the solution


Dominik about 8 yearsYou forgot to multiply by $a$.

jameselmore about 8 years@Dominik, thanks