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 \[d^{2}y/dx^{2} = (abc + 2fgh – af^{2} – bg^{2} – ch^{2})/(hx + by + f)^{3}.\]
