We may form a new function $$\phi”(x)$$ from $$\phi'(x)$$ just as we formed $$\phi'(x)$$ from $$\phi(x)$$. This function is called the second derivative or second differential coefficient of $$\phi(x)$$. The second derivative of $$y = \phi(x)$$ may also be written in any of the forms $D_{x}^{2}y,\quad \left(\frac{d}{dx}\right)^{2}y,\quad \frac{d^{2}y}{dx^{2}}.$

In exactly the same way we may define the $$n$$th derivative or $$n$$th differential coefficient of $$y = \phi(x)$$, which may be written in any of the forms $\phi^{(n)}(x),\quad D_{x}^{n}y,\quad \left(\frac{d}{dx}\right)^{n}y,\quad \frac{d^{n}y}{dx^{n}}.$ But it is only in a few cases that it is easy to write down a general formula for the $$n$$th differential coefficient of a given function. Some of these cases will be found in the examples which follow.

Example XLV

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

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

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 $\tfrac{1}{2}(n!) \{(1 – x)^{-n-1} + (-1)^{n}(1 + x)^{-n-1}\}.$

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 $(uv)_{n} = u_{n}v + \binom{n}{1}u_{n-1}v_{1} + \binom{n}{2}u_{n-2}v_{2} + \dots + \binom{n}{r}u_{n-r}v_{r} + \dots + uv_{n},$ 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{aligned} (uv)_{1} &= u_{1}v + uv_{1},\\ (uv)_{2} &= u_{2}v + 2u_{1}v_{1} + uv_{2},\end{aligned} and so on. It is obvious that by repeating this process we arrive at a formula of the type $(uv)_{n} = u_{n}v + a_{n, 1} u_{n-1} v_{1} + a_{n, 2} u_{n-2} v_{2} + \dots + a_{n, r} u_{n-r} v_{r} + \dots + uv_{n}.$

Let us assume that $$a_{n, r} = \dbinom{n}{r}$$ for $$r = 1$$, $$2$$, …, $$n – 1$$, and show that if this is so then $$a_{n+1, r} = \dbinom{n + 1}{r}$$ for $$r = 1$$, $$2$$, … $$n$$. It will then follow by the principle of mathematical induction that $$a_{n, r} = \dbinom{n}{r}$$ 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 $a_{n, r} + a_{n, r-1} = \binom{n}{r} + \binom{n}{r – 1} = \binom{n + 1}{r}.$ This establishes the theorem.

6. The $$n$$th derivative of $$x^{m}f(x)$$ is $\begin{gathered} \frac{m!}{(m – n)!} x^{m-n} f(x) + n \frac{m!}{(m – n + 1)!} x^{m-n+1} f'(x)\\ + \frac{n(n – 1)}{1\cdot2}\, \frac{m!}{(m – n + 2)!} x^{m-n+2} f”(x) + \dots,\end{gathered}$ 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 $\frac{a}{a^{2} + x^{2}} = \frac{1}{2i} \left(\frac{1}{x – ai} – \frac{1}{x + ai}\right), \quad \frac{x}{a^{2} + x^{2}} = \frac{1}{2} \left(\frac{1}{x – ai} + \frac{1}{x + ai}\right),$ 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 $$x + ai = \rho\operatorname{Cis}\theta$$ and $$x – ai = \rho\operatorname{Cis}(-\theta )$$, and so \begin{aligned} D_{x}^{n} \{a/(a^{2} + x^{2})\} &= \{(-1)^{n} n!/2i\} \rho^{-n-1} [\operatorname{Cis} \{(n + 1)\theta\} – \operatorname{Cis} \{-(n + 1)\theta\}]\\ &= (-1)^{n} n!\, (x^{2} + a^{2})^{-(n+1)/2} \sin \{(n + 1) \arctan(a/x)\}.\end{aligned} 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{aligned} D_{x}^{n} \{(\cos x)/x\} &= \{P_{n} \cos(x + \tfrac{1}{2}n\pi) + Q_{n} \sin(x + \tfrac{1}{2}n\pi)\}/x^{n+1},\\ D_{x}^{n} \{(\sin x)/x\} &= \{P_{n} \sin(x + \tfrac{1}{2}n\pi) – Q_{n} \cos(x + \tfrac{1}{2}n\pi)\}/x^{n+1},\end{aligned} where $$P_{n}$$ and $$Q_{n}$$ are polynomials in $$x$$ of degree $$n$$ and $$n-1$$ respectively.

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

14. If $$yz = 1$$ and $$y_{r} = (1/r!) D_{x}^{r}y$$, $$z_{s} = (1/s!) D_{x}^{s}z$$, then $\frac{1}{z^{3}} \begin{vmatrix} z & z_{1}& z_{2}\\ z_{1}& z_{2}& z_{3}\\ z_{2}& z_{3}& z_{4} \end{vmatrix} = \frac{1}{y^{2}} \begin{vmatrix} y_{2}& y_{3}\\ y_{3}& y_{4} \end{vmatrix}.$

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

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