120. Repeated differentiation

1. If $\varphi (x)=x^m$ then $${\varphi }^{(n)}(x)=m(m–1)\dots (m–n+1)x^{m-n}.$$ This result enables us to write down the $n$th derivative of any polynomial.

2. If $\varphi (x)=(ax+b{)}^m$ then $${\varphi }^{(n)}(x)=m(m–1)\dots (m–n+1)a^n(ax+b{)}^{m-n}.$$ In these two examples $m$ may have any rational value. If $m$ is a positive integer, and $n>m$, then ${\varphi }^{(n)}(x)=0$.

3. The formula $${\left(\frac{d}{dx}\right)}^n\frac{A}{(x–\alpha {)}^p}=(-1{)}^n\frac{p(p+1)\dots (p+n–1)A}{(x–\alpha {)}^{p+n}}$$ enables us to write down the $n$th derivative of any rational function expressed in the standard form as a sum of partial fractions.

4. Prove that the $n$th derivative of $1/(1–x^2)$ is $$\frac{1}{2}(n!)\{(1–x{)}^{-n-1}+(-1{)}^n(1+x{)}^{-n-1}\}.$$

5. Leibniz’ Theorem. If $y$ is a product $uv$, and we can form the first $n$ derivatives of $u$ and $v$, then we can form the $n$th derivative of $y$ by means of Leibniz’ Theorem, which gives the rule $$(uv{)}_n=u_{n}v+(\genfrac{}{}{0ex}{}{n}{1})u_{n-1}{v}_1+(\genfrac{}{}{0ex}{}{n}{2})u_{n-2}{v}_2+\cdots +(\genfrac{}{}{0ex}{}{n}{r})u_{n-r}{v}_r+\cdots +u{v}_n,$$ where suffixes indicate differentiations, so that $u_n$, for example, denotes the $n$th derivative of $u$. To prove the theorem we observe that $$\begin{array}{rl}(uv{)}_1& =u_{1}v+u{v}_1,\\ (uv{)}_2& =u_{2}v+2u_1{v}_1+u{v}_2,\end{array}$$ and so on. It is obvious that by repeating this process we arrive at a formula of the type $$(uv{)}_n=u_{n}v+a_{n,1}{u}_{n-1}{v}_1+a_{n,2}{u}_{n-2}{v}_2+\cdots +a_{n,r}{u}_{n-r}{v}_r+\cdots +u{v}_n.$$

Let us assume that $a_{n,r}={\displaystyle (\genfrac{}{}{0ex}{}{n}{r})}$ for $r=1$, $2$, …, $n–1$, and show that if this is so then $a_{n+1,r}={\displaystyle (\genfrac{}{}{0ex}{}{n+1}{r})}$ for $r=1$, $2$, … $n$. It will then follow by the principle of mathematical induction that $a_{n,r}={\displaystyle (\genfrac{}{}{0ex}{}{n}{r})}$ for all values of $n$ and $r$ in question.

When we form $(uv{)}_{n+1}$ by differentiating $(uv{)}_n$ it is clear that the coefficient of $u_{n+1-r}{v}_r$ is $$a_{n,r}+a_{n,r-1}=(\genfrac{}{}{0ex}{}{n}{r})+(\genfrac{}{}{0ex}{}{n}{r–1})=(\genfrac{}{}{0ex}{}{n+1}{r}).$$ This establishes the theorem.

6. The $n$th derivative of $x^{m}f(x)$ is $$\begin{array}{c}\frac{m!}{(m–n)!}{x}^{m-n}f(x)+n\frac{m!}{(m–n+1)!}{x}^{m-n+1}{f}^{\prime }(x)\\ +\frac{n(n–1)}{1\cdot 2}\frac{m!}{(m–n+2)!}{x}^{m-n+2}f”(x)+\dots ,\end{array}$$ the series being continued for $n+1$ terms or until it terminates.

7. Prove that $D_x^n\cos x=\cos (x+\frac{1}{2}n\pi )$, $D_x^n\sin x=\sin (x+\frac{1}{2}n\pi )$

8. If $y=A\cos mx+B\sin mx$ then $D_x^{2}y+m^{2}y=0$. And if $$y=A\cos mx+B\sin mx+P_n(x),$$ where $P_n(x)$ is a polynomial of degree $n$, then $D_x^{n+3}y+m^2{D}_x^{n+1}y=0$.

9. If $x^2{D}_x^{2}y+x{D}_{x}y+y=0$ then $$x^2{D}_x^{n+2}y+(2n+1)x{D}_x^{n+1}y+(n^2+1)D_x^{n}y=0.$$

[Differentiate $n$ times by Leibniz’ Theorem.]

10. If $U_n$ denotes the $n$th derivative of $(Lx+M)/(x^2–2Bx+C)$, then $$\frac{x^2–2Bx+C}{(n+1)(n+2)}{U}_{n+2}+\frac{2(x–B)}{n+1}{U}_{n+1}+U_n=0.$$

[First obtain the equation when $n=0$; then differentiate $n$ times by Leibniz’ Theorem.]

11. The $n$th derivatives of $a/(a^2+x^2)$ and $x/(a^2+x^2)$. Since $$\frac{a}{a^2+x^2}=\frac{1}{2i}\left(\frac{1}{x–ai}–\frac{1}{x+ai}\right),\frac{x}{a^2+x^2}=\frac{1}{2}(\frac{1}{x–ai}+\frac{1}{x+ai}),$$ we have $$D_x^n\left(\frac{a}{a^2+x^2}\right)=\frac{(-1{)}^{n}n!}{2i}\left\{\frac{1}{(x–ai{)}^{n+1}}–\frac{1}{(x+ai{)}^{n+1}}\right\},$$ and a similar formula for $D_x^n\{x/(a^2+x^2)\}$. If $\rho =\sqrt{x^2+a^2}$, and $\theta$ is the numerically smallest angle whose cosine and sine are $x/\rho$ and $a/\rho$, then $x+ai=\rho \mathrm{Cis}\theta$ and $x–ai=\rho \mathrm{Cis}(-\theta )$, and so $$\begin{array}{rl}{D}_x^n\{a/(a^2+x^2)\}& =\{(-1{)}^{n}n!/2i\}{\rho }^{-n-1}[\mathrm{Cis}\{(n+1)\theta \}–\mathrm{Cis}\{-(n+1)\theta \}]\\ & =(-1{)}^{n}n!(x^2+a^2{)}^{-(n+1)/2}\sin \{(n+1)\arctan (a/x)\}.\end{array}$$ Similarly $$D_x^n\{x/(a^2+x^2)\}=(-1{)}^{n}n!(x^2+a^2{)}^{-(n+1)/2}\cos \{(n+1)\arctan (a/x)\}.$$

12. Prove that $$\begin{array}{rl}{D}_x^n\{(\cos x)/x\}& =\{P_n\cos (x+\frac{1}{2}n\pi )+Q_n\sin (x+\frac{1}{2}n\pi )\}/x^{n+1},\\ D_x^n\{(\sin x)/x\}& =\{P_n\sin (x+\frac{1}{2}n\pi )–Q_n\cos (x+\frac{1}{2}n\pi )\}/x^{n+1},\end{array}$$ where $P_n$ and $Q_n$ are polynomials in $x$ of degree $n$ and $n-1$ respectively.

13. Establish the formulae $$\begin{array}{c}\frac{dx}{dy}=1/(\frac{dy}{dx}),\frac{d^{2}x}{d{y}^2}=-\frac{d^{2}y}{d{x}^2}/(\frac{dy}{dx}{)}^3,\\ \frac{d^{3}x}{d{y}^3}=-\{\frac{d^{3}y}{d{x}^3}\frac{dy}{dx}–3(\frac{d^{2}y}{d{x}^2})\}/(\frac{dy}{dx}{)}^5.\end{array}$$

14. If $yz=1$ and $y_r=(1/r!)D_x^{r}y$, $z_s=(1/s!)D_x^{s}z$, then $$\frac{1}{z^3}\left|\begin{array}{ccc}z& z_1& z_2\\ z_1& z_2& z_3\\ z_2& z_3& z_4\end{array}\right|=\frac{1}{y^2}\left|\begin{array}{cc}{y}_2& y_3\\ y_3& y_4\end{array}\right|.$$

15. If $$W(y,z,u)=\left|\begin{array}{ccc}y& z& u\\ y^{\prime }& z^{\prime }& u^{\prime }\\ y”& z”& u”\end{array}\right|,$$ dashes denoting differentiations with respect to $x$, then $$W(y,z,u)=y^{3}W(1,\frac{z}{y},\frac{u}{y}).$$

16. If $$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0,$$ then $$dy/dx=-(ax+hy+g)/(hx+by+f)$$ and $$d^{2}y/d{x}^2=(abc+2fgh–a{f}^2–b{g}^2–c{h}^2)/(hx+by+f{)}^3.$$