Show that $y^{(n)}=\frac{m!a^n}{(m-n)!}(ax+b)^{m-n}$

Question

If $y=(ax+b)^m,$ then show that $$y^{(n)}=\frac{m!a^n}{(m-n)!}(ax+b)^{m-n},$$

where $y^{(n)}$ is the $n$-th derivative of $y$ with respect to $x$.

They say $$y^{(n)}=\frac{m(m-1)(m-2)...(m-n+1)(m-n)!}{(m-n)!} (ax+b)^{m-n} \cdot a^n$$

$$=\frac{m!a^n}{(m-n)!} (ax+b)^{m-n} \cdot a^n$$

I couldn't understand second line. Usually, I didn't solve many problems with factorial. That's why I don't have much more knowledge of factorial. I didn't understand what it ($(m-n)!$) was converted in second line. Where did $a$ come from?

Answer 1

We start right from the beginning to better see what's going on. Differentiating $y(x)=(ax+b)^m$ we obtain \begin{align*} y(x)&=(ax+b)^m\\ y^{\prime}(x)&=m(ax+b)^{m-1}a\tag{1}\\ y^{\prime\prime}(x)&=m(m-1)(ax+b)^{m-2}a^2\tag{2}\\ y^{(3)}(x)&=m(m-1)(m-2)(ax+b)^{m-3}a^3\\ &\vdots\\ y^{(n)}(x)&=\color{blue}{m(m-1)\cdots (m-n+1)}(ax+b)^{m-n+1}a^n\tag{3}\\ \end{align*}

Comment:

• In (1) we derivate with respect to $x$ noting that the inner derivative of $ax+b$ gives a factor $a$.

• In (2) down to (3) we derivate successively and see two patterns emerging: A factor $m(m-1)\cdots (m-n+1)$ and according to $n$-times doing inner derivation a factor $a^n$.

We recall the definition of $m!=m\cdot (m-1)\cdot (m-2)\cdots 3\cdot 2\cdot 1$. We consider the inner factor $m-n+1$ of $m!$ and group $m!$ accordingly to get \begin{align*} m!&=m\cdot (m-1)\cdot (m-2)\cdots 3\cdot 2\cdot 1\\ &=m\cdot(m-1)\cdot(m-2)\cdots (m-n+1)\\ &\qquad\qquad\cdot\color{blue}{\left((m-n)\cdot (m-n-1)\cdots 3\cdot 2\cdot1\right)}\\ &=m\cdot(m-1)\cdot(m-2)\cdots (m-n+1)\cdot \color{blue}{(m-n)!}\tag{4} \end{align*}

Expanding (3) with $(m-n)!$ we obtain according to (4) \begin{align*} \color{blue}{y^{(n)}(x)}&=m(m-1)\cdots (m-n)(ax+b)^{m-n+1}a^n\\ &=\frac{m(m-1)\cdots (m-n+1)(ax+b)^{m-n}a^n(m-n)!}{(m-n)!}\\ &\;\;\color{blue}{=\frac{m!a^n(ax+b)^{m-n+1}}{(m-n)!}} \end{align*} and the claim follows.