Derivatives of Higher Order
Let us try the effect of repeating several times over the operation of differentiating a function (see the concept of a function). Begin with a concrete case.
Let \(y = x^5\). \[\begin{align} &\text{First differentiation, } &&5x^4. && \\ &\text{Second differentiation, } &&5 \times 4x^3 &&= 20x^3. \\ &\text{Third differentiation, } &&5 \times 4 \times 3x^2 &&= 60x^2. \\ &\text{Fourth differentiation, } &&5 \times 4 \times 3 \times 2x &&= 120x. \\ &\text{Fifth differentiation, } &&5 \times 4 \times 3 \times 2 \times 1 &&= 120. \\ &\text{Sixth differentiation, } && &&= 0. \end{align}\]
There is a certain notation, with which we are already acquainted, used by some writers, that is very convenient. This is to employ the general symbol \(f(x)\) for any function of \(x\). Here the symbol \(f(~)\) is read as “function of,” without saying what particular function is meant. So the statement \(y=f(x)\) merely tells us that \(y\) is a function of \(x\), it may be \(x^2\) or \(ax^n\), or \(\cos x\) or any other complicated function of \(x\).
The corresponding symbol for the derivative is \(f^{\prime}(x)\), which is simpler to write than \(\dfrac{dy}{dx}\). This is called the derivative of \(\boldsymbol{y}\) with respect to \(\boldsymbol{x}\), the derivative of the function \(\boldsymbol{f}\), or simply the derivative function. Instead of \(\dfrac{dy}{dx}\) or \(f^\prime(x)\), we sometimes simply write \(y^\prime\).
Suppose we differentiate over again, we shall get the second derivative of \(\boldsymbol{f}\) or the second derivative of \(\boldsymbol{y}\) with respect to \(\boldsymbol{x}\), which is denoted by \(f^{\prime\prime}(x)\) or \(y^{\prime\prime}\); and so on.
Now let us generalize.
Let \(y = f(x) = x^n\). \[\begin{align} &\text{First differentiation,} &&y^\prime=f^\prime(x) = nx^{n-1}. \\ &\text{Second differentiation,} &&y^{\prime\prime}=f^{\prime\prime}(x) = n(n-1)x^{n-2}. \\ &\text{Third differentiation,} &&y^{\prime\prime\prime}=f^{\prime\prime\prime}(x) = n(n-1)(n-2)x^{n-3}. \\ &\text{Fourth differentiation,} &&y^{\prime\prime\prime\prime}=f^{\prime\prime\prime\prime}(x) = n(n-1)(n-2)(n-3)x^{n-4}. \\ &&\vdots \end{align}\]
In general, after differentiating the original function \(n\) times, we obtain the \(n\)-th derivative of \(f\) or \(y\) with respect to \(x\), also known as the derivative of order \(n\). When the differentiation order reaches four or more, rather than repeatedly using accents (also known as primes), a more streamlined approach is often adopted. The order of differentiation is denoted using parentheses, with the derivative order presented as a superscript to \(f\) or \(y\). This notation is not only clearer but also helps reduce the risk of miscounting the number of primes. For instance, we often write \(y^{(4)}\) or \(f^{(4)}(x)\) instead of \(y^{\prime\prime\prime\prime}\) and \(f^{\prime\prime\prime\prime}(x)\).
There is another way of indicating successive differentiations. For, \[\begin{align} &\text{if the original function is } &&y = f(x); \\ &\text{once differentiating gives } &&\frac{dy}{dx} = f^{\prime}(x); \\ &\text{twice differentiating gives } &&\frac{d\left(\dfrac{dy}{dx}\right)}{dx} = f^{\prime\prime}(x); \end{align}\] and this is more conveniently written as \(\dfrac{d^2y}{(dx)^2}\), or more usually \(\dfrac{d^2y}{dx^2}\). Similarly, we may write as the result of differentiating three times, \(\dfrac{d^3y}{dx^3} = f^{\prime\prime\prime}(x)\).
How to Read the Symbols for Derivatives
| \(f^\prime(x)\) | eff prime of eks |
| \(f^{\prime\prime} (x)\) | eff double prime of eks |
| \(f^{\prime\prime\prime}(x)\) | eff triple prime of eks |
| \(f^{(n)}(x)\) | eff super en of eks (or the en-th derivative of eff of eks) |
| \(y^\prime\) | wy prime |
| \(y^{\prime\prime}\) | wy double prime |
| \(y^{\prime\prime\prime}\) | wy triple prime |
| \(y^{(n)}\) | why super en (or the en-th derivative of wy) |
| \(\dfrac{dy}{dx}\) | dee wy over dee eks |
| \(\dfrac{d^2 y}{dx^2}\) | dee squared wy over dee eks squared |
Examples
Now let us try \(y = f(x) = 7x^4 + 3.5x^3 - \frac{1}{2}x^2 + x - 2\). \[\begin{align} \frac{dy}{dx} &= f^{\prime}(x) = 28x^3 + 10.5x^2 - x + 1, \\ \frac{d^2y}{dx^2} &= f^{\prime\prime}(x) = 84x^2 + 21x - 1, \\ \frac{d^3y}{dx^3} &= f^{\prime\prime\prime}(x) = 168x + 21, \\ \frac{d^4y}{dx^4} &= f^{(4)}(x) = 168, \\ \frac{d^5y}{dx^5} &= f^{(5)}(x) = 0. \end{align}\]
In a similar manner if \(y = \phi(x) = 3x(x^2 - 4)\), \[\begin{align} \phi^\prime(x) &= \frac{dy}{dx} = 3\bigl[x \times 2x + (x^2 - 4) \times 1\bigr] = 3(3x^2 - 4), \\ \phi^{\prime\prime}(x) &= \frac{d^2y}{dx^2} = 3 \times 6x = 18x, \\ \phi^{\prime\prime\prime}(x) &= \frac{d^3y}{dx^3} = 18, \\ \phi^{(4)}(x) &= \frac{d^4y}{dx^4} = 0. \end{align}\]
Exercises
Find \(\dfrac{dy}{dx}\) and \(\dfrac{d^2y}{dx^2}\) for the following expressions:
Exercise 7.1. \(y = 17x + 12x^2\).
Answer
\(17 + 24x\);\(24\).
Solution
\[\begin{align} & y=17 x+12 x^{2} \\ & \frac{d y}{d x}=17+24 x \\ & \frac{d^{2} y}{d x^{2}}=24 \end{align}\]
Exercise 7.2. \(y = \dfrac{x^2 + a}{x + a}\).
Answer
\(\dfrac{x^2 + 2ax - a}{(x + a)^2}\);\(\dfrac{2a(a + 1)}{(x + a)^3}\).
Solution
\[y=\frac{x^{2}+a}{x+a}\] Using the Quotient Rule
\[\begin{align} \frac{d y}{d x} & =\frac{2 x(x+a)-\left(x^{2}+a\right)}{(x+a)^{2}} \\ & =\frac{x^{2}+2 a x-a}{(x+a)^{2}} \end{align}\]
To find \(\frac{d^{2} y}{d x^{2}}\), we use the Quotient Rule again.
\[\begin{align} \frac{d^{2} y}{d x^{2}}&=\frac{(2 x+2 a)(x+a)^{2}-\frac{d\left(x^{2}+2 a x+a\right)^{2}}{d x}\left(x^{2}+2 a x-a\right)}{(x+a)^{4}}\\ & =\frac{2(x+a)^{3}-(2 x+2 a)\left(x^{2}+2 a x-a\right)}{(x+a)^{4}} \\ & =\frac{2(x+a)\left[(x+a)^{2}-\left(x^{2}+2 a x-a\right)\right]}{(x+a)^{4}}\\ & =\frac{2\left[x^{2}+2 a x+a^{2}-x^{2}-2 a x+a\right]}{(x+a)^{3}} \\ & =\frac{2\left(a^{2}+a\right)}{(x+a)^{3}} \\ & =\frac{2 a(a+1)}{(x+a)^{3}} \end{align}\]
Exercise 7.3. \(y = 1 + \dfrac{x}{1} + \dfrac{x^2}{1\times2} + \dfrac{x^3}{1\times2\times3} + \dfrac{x^4}{1\times2\times3\times4}\).
Answer
\(1 + x + \dfrac{x^2}{1 \times 2} + \dfrac{x^3}{1 \times 2 \times 3}\);\(1 + x + \dfrac{x^2}{1 \times 2}\).
Solution
\[\begin{align} y & =1+\frac{x}{1}+\frac{x^{2}}{1 \times 2}+\frac{x^{3}}{1 \times 2 \times 3}+\frac{x^{4}}{1 \times 2 \times 3 \times 4}. \\ \frac{d y}{d x} & =1+\frac{x}{1}+\frac{x^{2}}{1 \times 2}+\frac{x^{3}}{1 \times 2 \times 3}. \\ \frac{d^{2} y}{d x} & =1+\frac{x}{1}+\frac{x^{2}}{1 \times 2}. \end{align}\]
Exercise 7.4. Find the 2nd and 3rd derivatives in the Exercises of Chapter 6, No. 1 to No. 7:
Expressions:
- First Exercise:
- \(u = 1 + x + \dfrac{x^2}{1 \times 2} + \dfrac{x^3}{1 \times 2 \times 3} + \dotsb\).
- \(y = ax^2 + bx + c\).
- \(y = (x + a)^2\).
- \(y = (x + a)^3\).
- \(w = at - \frac{1}{2}bt^2\).
- \(y = (x + \sqrt{-1}) \times (x - \sqrt{-1}).\)
- \(y = (197x - 34x^2) \times (7 + 22x - 83x^3).\)
- \(x = (y + 3) \times (y + 5)\).
- \(y = 1.3709x \times (112.6 + 45.202x^2)\).
- \(y = \dfrac{2x + 3}{3x + 2}\).
and in Example 6.4 to Example 6.10:
Example 6.4: \(y = \dfrac{a}{b^2} x^3 - \dfrac{a^2}{b} x + \dfrac{a^2}{b^2}\).
Example 6.5: \(y = 2a\sqrt{bx^3} - \dfrac{3b \sqrt[3]{a}}{x} - 2\sqrt{ab}\)
Example 6.6: \(z = 1.8 \sqrt[3]{\dfrac{1}{\theta^2}} - \dfrac{4.4}{\sqrt[5]{\theta}} - 27\).
Example 6.7: \(v = (3 t^2 - 1.2 t + 1)^3\)
Example 6.8: \(y = (2x - 3)(x + 1)^2\).
Example 6.9: \(y = 0.5 x^3(x-3)\).
Example 6.10: \(w = \left(\theta + \dfrac{1}{\theta}\right) \left(\sqrt{\theta} + \dfrac{1}{\sqrt{\theta}}\right)\).
Answer
(Exercises of Chapter 6):
(1) \(\dfrac{d^2 y}{dx^2} = \dfrac{d^3 y}{dx^3} = 1 + x + \frac{1}{2}x^2 + \frac{1}{6} x^3 + \ldots\).
(2) \(2a\), \(0\).
(3) \(2\), \(0\).
(4) \(6x + 6a\), \(6\).
\(-b\), \(0\).
\(2\), \(0\).
\(\begin{gathered}[t] 56440x^3 - 196212x^2 - 4488x + 8192. \\ 169320x^2 - 392424x - 4488. \end{gathered}\)
\(2\), \(0\).
\(371.80453x\), \(371.80453\).
\(\dfrac{30}{(3x + 2)^3}\),\(-\dfrac{270}{(3x + 2)^4}\).
Example 6.4: \(\dfrac{6a}{b^2} x\),\(\dfrac{6a}{b^2}\).
Example 6.5: \(\dfrac{3a \sqrt{b}} {2 \sqrt{x}} - \dfrac{6b \sqrt[3]{a}}{x^3}\), \(\dfrac{18b \sqrt[3]{a}}{x^4} - \dfrac{3a \sqrt{b}}{4 \sqrt{x^3}}\).
Example 6.6: \(\dfrac{2}{\sqrt[3]{\theta^8}} - \dfrac{1.056}{\sqrt[5]{\theta^{11}}}\), \(\dfrac{2.3232}{\sqrt[5]{\theta^{16}}} - \dfrac{16}{3 \sqrt[3]{\theta^{11}}}\).
Example 6.7: \(810t^4 - 648t^3 + 479.52t^2 - 139.968t + 26.64.\), \(3240t^3 - 1944t^2 + 959.04t - 139.968.\)
Example 6.8: \(12x + 2\), \(12\).
Example 6.9: \(6x^2 - 9x\),\(12x - 9\).
Example 6.10: \[\begin{align} &\dfrac{3}{4} \left(\dfrac{1}{\sqrt{\theta}} + \dfrac{1}{\sqrt{\theta^5}}\right) +\dfrac{1}{4} \left(\dfrac{15}{\sqrt{\theta^7}} - \dfrac{1}{\sqrt{\theta^3}}\right). \\ &\dfrac{3}{8} \left(\dfrac{1}{\sqrt{\theta^5}} - \dfrac{1}{\sqrt{\theta^3}}\right) -\dfrac{15}{8}\left(\dfrac{7}{\sqrt{\theta^9}} + \dfrac{1}{\sqrt{\theta^7}}\right). \end{align}\]
Solution
(1)
(a) We learned that
\[\frac{d u}{d x}=u\]
Therefore,
\[\frac{d^{2} u}{d x^{2}}=\frac{d\left(\frac{d u}{d x}\right)}{d x}=\frac{d u}{d x}=u\]
and
\[\frac{d^{3} u}{d x^{3}}=\frac{d\left(\frac{d^{2} u}{d x^{2}}\right)}{d x}=\frac{d u}{d x}=u .\]
(b) Since \[\frac{d y}{d x}=2 a x+b\] then \[\begin{align} & \frac{d^{2} y}{d x^{2}}=2 a \\ & \frac{d^{3} y}{d x^{3}}=0 \end{align}\]
(c) Since \(\frac{d y}{d x}=2(x+a)=2 x+2 a\)
\[\begin{align} & \frac{d^{2} y}{d x^{2}}=2 \\ & \frac{d^{3} y}{d x^{3}}=0 \end{align}\]
(d) Since
\[\begin{align} & \frac{d y}{d x}=3(x+a)^{2}=3\left(x^{2}+2 a x+a^{2}\right) \\ & \frac{d^{2} y}{d x^{2}}=6 x+6 a=6(x+a) \\ & \frac{d^{3} y}{d x^{3}}=6 \end{align}\]
(2) Since \(\frac{d w}{d t}=a-b t\)
\[\begin{align} & \frac{d^{2} w}{d t^{2}}=-b \\ & \frac{d^{3} w}{d t^{3}}=0 \end{align}\]
(3) Since \[\frac{d y}{d x}=2 x,\] \[\begin{align} & \frac{d^{2} y}{d x^{2}}=2 \quad \text { and } \quad \frac{d^{3} y}{d x^{3}}=0 \end{align}\]
(4) Since \(\frac{d y}{d x}=14110 x^{4}-65404 x^{3}-22404 x^{2}+8192 x+1379\),
\[\begin{align} \frac{d^{2} y}{d x^{2}} & =56440 x^{3}-196212 x^{2}-44808 x+8192 \\ \frac{d^{3} y}{d x^{3}} & =3 \times 56440 x^{2}-2 \times 196212 x-44808 \\ & =169320 x^{2}-392424 x-44808 \end{align}\]
(5) Since \(\frac{d x}{d y}=2 y+5\)
\[ \frac{d^2 x}{d y^{2}}=2 \quad \text { and }\quad \frac{d^{3} x}{d y^{3}}=0 \]
(6) Since \(\dfrac{d y}{d x}=185.9022654 x^2+154.36334\)
\[\begin{align} & \frac{d^2 y}{d x^2}=2 \times 185.9022654 x=371.8045308 x \\ & \frac{d^3 y}{d x^3}=371.8045308 \end{align}\]
(7) Since \(\dfrac{d y}{d x}=-\dfrac{5}{(3 x+2)^{2}}\), \[\begin{align} \frac{d^{2} y}{d x^{2}} & =-\frac{0 \times(3 x+2)^{2}-5 \frac{d\left[(3 x+2)^{2}\right]}{d x}}{(3 x+2)^{4}} \\ & =\frac{5 \frac{d\left[9 x^{2}+12 x+4\right]}{d x}}{(3 x+2)^{4}} \\ & =\frac{5(18 x+12)}{(3 x+2)^{4}} \\ & =\frac{30(3 x+2)}{(3 x+2)^{4}} \\ & =\frac{30}{(3 x+2)^{3}} \\ \frac{d^{3} y}{d x^{3}} & =\frac{-30 \frac{d\left[(3 x+2)^{3}\right]}{d x}}{(3 x+2)^{6}} \\ & =-\frac{30 \frac{d\left(27 x^{3}+54 x^{2}+36 x+8\right)}{d x}}{(3 x+2)^{6}} \\ & =\frac{30\left(81 x^{2}+108 x+36\right)}{(3 x+2)^{6}} \end{align}\]
\[\begin{align} & =\frac{30 \times 9\left(9 x^{2}+12 x+4\right)}{(3 x+2)^{6}} \\ & =\frac{270(3 x+2)^{2}}{(3 x+2)^{6}} \\ & =\frac{270}{(3 x+2)^{5}} \end{align}\]
Example 19. Since \(\dfrac{d y}{d x}=\dfrac{3 a}{b^{2}} x^{2}-\dfrac{a^{2}}{b}\)
\[\frac{d^{2} y}{d x^{2}}=\frac{6 a}{b^{2}} x \text { and } \frac{d^{3} y}{d x^{3}}=\frac{6 a}{b^{2}}\]
Example 20. Since \(\dfrac{d y}{d x}=3 a \sqrt{b x}+\dfrac{3 b \sqrt[3]{a}}{x^{2}}\). We may rewrite it as \[\frac{d y}{d x}=3 a \sqrt{b} x^{\frac{1}{2}}+3 b \sqrt[3]{a} x^{-2}\] Therefore \[\begin{align} \frac{d^2 y}{d x^2} & =\frac{1}{2} 3 a \sqrt{b} x^{-\frac{1}{2}}-6 b \sqrt[3]{a} x^{-3} \\ & =\frac{3}{2} a \sqrt{\frac{b}{x}}-\frac{6 b \sqrt[3]{a}}{x^3} \end{align}\] and
\[\begin{align} \frac{d^{3} y}{d x^{3}} & =-\frac{1}{4} 3 a \sqrt{b} x^{-\frac{3}{2}}+18 b \sqrt[3]{a} x^{-4} \\ & =-\frac{3 a \sqrt{b}}{4 \sqrt{x^{3}}}+\frac{18 b \sqrt[4]{a}}{x^{4}} \end{align}\]
Example 21. Since \(\dfrac{d z}{d \theta}=-1.2 \theta^{-\frac{5}{3}}+0.88 \theta^{-\frac{6}{5}}\)
\[\begin{align} \frac{d^{2} z}{d \theta^{2}} & =2 \theta^{-\frac{8}{3}}-1.056 \theta^{-\frac{11}{5}} \\ & =\frac{2}{\sqrt[3]{\theta^{8}}}-\frac{1.056}{\sqrt[5]{\theta^{11}}} \\ \frac{d^{3} z}{d \theta^{3}} & =-\frac{16}{3} \theta^{-\frac{11}{3}}+2.3232 \theta^{-\frac{16}{5}} \\ & =-\frac{16}{\sqrt[3]{\theta^{11}}}+\frac{2.3232}{\sqrt[5]{\theta^{16}}} \end{align}\]
Example 22. Since \(\dfrac{d v}{d t}=162 t^{5}-162 t^{4}+159.84 t^{3}-69.984 t^{2}+26.64 t\)
\[\begin{align} & \frac{d^{2} v}{d t^{2}}=810 t^{4}-648 t^{3}+479.52 t^{2}-139.968 t+26.64 \\ & \frac{d^{3} v}{d t^{3}}=3240 t^{3}-1944 t^{2}+959.04 t-139.968 \end{align}\]
Example 23. Since \(\frac{d y}{d x}=2(x+1)(3 x-2)\)
\[\frac{d^{2} y}{d x^{2}}=2(3 x-2)+6(x+1)=12 x+2\]
\[\frac{d^{3} y}{d x^{3}}=12\]
Example 24. Since \(\frac{d y}{d x}=2 x^{3}-4.5 x^{2}\)
\[\begin{align} & \frac{d^{2} y}{d x^{2}}=6 x^{2}-9 x \\ & \frac{d^{3} y}{d x^{3}}=12 x-9 \end{align}\]
Example 25. Since \(\dfrac{d \omega}{d \theta}=\frac{3}{2}\left(\sqrt{\theta}-\frac{1}{\sqrt{\theta^{5}}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{\theta}}-\frac{1}{\sqrt{\theta^{3}}}\right)\), we can rewrite it as
\[\frac{d w}{d \theta}=\frac{3}{2}\left(\theta^{\frac{1}{2}}-\theta^{-\frac{5}{2}}\right)+\frac{1}{2}\left(\theta^{-\frac{1}{2}}-\theta^{-\frac{3}{2}}\right)\] Therefore
\[\begin{align} \frac{d^{2} w}{d \theta^{2}} & =\frac{3}{2}\left(\frac{1}{2} \theta^{-\frac{1}{2}}+\frac{5}{2} \theta^{-\frac{7}{2}}\right)+\frac{1}{2}\left(-\frac{1}{2} \theta^{-\frac{3}{2}}+\frac{3}{2} \theta^{-\frac{5}{2}}\right) \\ & =\frac{3}{4} \theta^{-\frac{1}{2}}+\frac{15}{4} \theta^{-\frac{7}{2}}-\frac{1}{4} \theta^{-\frac{3}{2}}+\frac{3}{4} \theta^{-\frac{5}{2}} \\ & =\frac{3}{4}\left(\theta^{-\frac{1}{2}}+\theta^{-\frac{5}{2}}\right)+\frac{1}{4}\left(15 \theta^{-\frac{7}{2}}-\theta^{-\frac{3}{2}}\right) \\ & =\frac{3}{4}\left(\frac{1}{\sqrt{\theta}}+\frac{1}{\sqrt{\theta^{5}}}\right)+\frac{1}{4}\left(\frac{15}{\sqrt{\theta^{7}}}-\frac{1}{\sqrt{\theta^{3}}}\right) \end{align}\]
\[\begin{align} \frac{d^{3} w}{d \theta^{3}} & =\frac{3}{4}\left(-\frac{1}{2} \theta^{\frac{2}{2}}-\frac{5}{2} \theta^{2}\right)+\frac{1}{4}\left(-\frac{105}{2} \theta^{-\frac{2}{2}}+\frac{3}{2} \theta^{-\frac{1}{2}}\right) \\ & =\frac{3}{8}\left(\theta^{-\frac{5}{2}}-\theta^{-\frac{3}{2}}\right)-\frac{15}{8}\left(7 \theta^{-\frac{9}{2}}+\theta^{-\frac{7}{2}}\right) \\ & =\frac{3}{8}\left(\frac{1}{\sqrt{\theta^{5}}}-\frac{1}{\sqrt{\theta^{3}}}\right)-\frac{15}{8}\left(\frac{7}{\sqrt{\theta^{9}}}+\frac{1}{\sqrt{\theta^{7}}}\right) \end{align}\]