Chapter 5 Higher Order Derivatives-I

Also called as successive derivatives.

Formulas for some typical functions

Table 1
Function	\(\text{n}^{th}\) derivative	Remarks
\(y=x^m\)	\(y_n =\frac{m!}{(m-n)!} x^{m-n}\)	\(m\) is a positive integer
\(y=x^m\)	\(y_n = n!\)	if \(m=n\) be a positive integer
\(y=(ax+b)^m\)	\(y_n = \frac{m!}{(m-n)!} a^n(ax+b)^{m-n}\)	\(m\) is a positive integer, \(n<m\)
\(y=\log x\)	\(y_n = \frac{(-1)^{n-1}(n-1)!}{x^n}\)
\(y=\log (ax+b)\)	\(y_n = \frac{(-1)^{n-1}(n-1)! a^n}{(ax+b)^n}\)
\(y=\frac{1}{x}\)	\(y_n = \frac{(-1)^n n!}{x^{n+1}}\)
\(y=\frac{1}{ax+b}\)	\(y_n = \frac{(-1)^n n! a^n}{(ax + b)^{n+1}}\)
\(y=a^{mx}\)	\(y_n = m^n a^{mx}(\log a)^n\)
\(y=e^{ax}\)	\(y_n = a^n e^{ax}\)
\(y=a^x\)	\(y_n = a^x (\log_e a)^n\)
\(y = \sin(ax+b)\)	\(y_n = a^n \sin (ax+b+\frac{n\pi}{2})\)
\(y = \cos (ax+b)\)	\(y_n=a^n \cos(ax+b + \frac{n\pi}{2})\)

Table 2
Function	\(\text{n}^{th}\) derivative	Remarks
\(y=e^{ax}\sin bx\)	\(y_n = (a^2 + b^2)^{n/2} e^{ax} \sin \left( bx + n \tan^{-1} \frac{b}{a}\right)\)
\(y=e^{ax}\cos bx\)	\(y_n = (a^2 + b^2)^{n/2} e^{ax} \cos \left( bx + n \tan^{-1} \frac{b}{a}\right)\)
\(y=\frac{1}{x^2-a^2}\)	\(y_n = \frac{(-1)^n n!}{2a}\left[\frac{1}{(x-a)^{n+1}} - \frac{1}{(x+a)^{n+1}} \right]\)
\(y=\frac{1}{x^2 + a^2}\)	\(y_n = \frac{(-1)^n n!}{a^{n+2}}\sin^{n+1}\theta \sin (n+1)\theta\)	where \(\theta = \tan^{-1}\left(\frac{a}{x}\right)\)
\(y=\tan^{-1}x\)	\(y_n = (-1)^{n-1} (n-1)! \sin^{n}\theta \sin n \theta\)	where \(\theta = \tan^{-1}\left(\frac{1}{x}\right)\)

5.1 Exercise 3 (i)

5.1.1 Question 1

Find \(y_n\) of the following:

\(\boldsymbol{y=(a-bx)^m}\)

\[\begin{equation*} \begin{split} y_1 &= -m (a-bx)^{m-1}b\\ y_2 &= m(m-1)(a-bx)^{m-2}b^2\\ y_3 &= -m(m-1)(m-2)(a-bx)^{m-3}b^3\\ & \vdots \\ y_n &= (-1)^n m(m-1)(m-2) \ldots (m-n+1)b^n(a-bx)^{m-n} \end{split} \end{equation*}\]

\(\boldsymbol{y=\frac{1}{a-x}}\)

\[\begin{equation*} \begin{split} y_1 &= -1(a-x)^{-2}\times -1 = (a-x)^{-2}\\ y_2 &= -2(a-x)^{-3}\times -1 = 2(a-x)^{-3}\\ y_3 &= -6 (a-x)^{-4} \times -1 = 6 (a-x)^{-4}\\ & \vdots \\ y_n &= \frac{n!}{(a-x)^{n+1}} \end{split} \end{equation*}\]

\(\boldsymbol{y=x^{2n}}\)

\[\begin{equation*} \begin{split} y_1 &= 2n x^{2n-1}\\ y_2 &= 2n(2n-1)x^{2n-2}\\ y_3 &= 2n(2n-1)(2n-2)x^{2n-3}\\ & \vdots \\ y_n &= 2n (2n-1)(2n-2)(2n-3)(2n-4) \ldots (2n-n+1) x^{2n-n}\\ &= 2n (2n-1)(2n-2) (2n-3)(2n-4) \ldots (n+1) x^n\\ &= \frac{2n (2n-1)(2n-2)(2n-3)(2n-4)\ldots (n+1)n (n-1) (n-2) (n-3)\ldots 3.2.1}{n (n-1) (n-2) (n-3)\ldots 3.2.1}x^n\\ \end{split} \end{equation*}\]

Separating even and odd factors, we write

\[\begin{equation*} \begin{split} y_n &= \frac{[2n(2n-2)(2n-4)(2n-8)\ldots 6.4.2][(2n-1)(2n-3)\ldots 5.3.1]}{n!} x^n\\ y_n &= \frac{2^n[n(n-1)(n-2)\ldots 3.2.1][1.3.5\ldots (2n-3)(2n-1)]}{n!} x^n\\ y_n &= 2^n[1.3.5\ldots (2n-3)(2n-1)]x^n \end{split} \end{equation*}\]

\(\boldsymbol{y=\sqrt{x}}\)

\[\begin{equation*} \begin{split} y_1 &= \frac{1}{2}x^{-1/2}\\ y_2 &= -\frac{1}{2\times 2}x^{-3/2}\\ y_3 &= \frac{1 \times 3}{2 \times 2 \times 2} x^{-5/2}\\ y_4 &= -\frac{1 \times 3 \times 5}{2 \times 2 \times 2 \times 2} x^{-7/2}\\ y_5 &= \frac{1 \times 3 \times 5 \times 7}{2 \times 2 \times 2 \times 2 \times 2} x^{-9/2}\\ & \vdots \\ y_n &= (-1)^{n-1} \frac{1.3.5.7.9 \ldots (2n-3)}{2^n x^{n-1/2}} \end{split} \end{equation*}\]

\(\boldsymbol{y=\frac{1}{\sqrt{x}}}\)

\[\begin{equation*} \begin{split} y_1 &= -\frac{1}{2} x^{-3/2}\\ y_2 &= \frac{1 \times 3}{2 \times 2} x^{-5/2}\\ y_3 &= - \frac{1 \times 3 \times 5}{2 \times 2 \times 2} x^{-7/2}\\ y_4 &= \frac{1 \times 3 \times 5 \times 7}{2 \times 2 \times 2 \times 2} x^{-9/2}\\ & \vdots \\ y_n &= (-1)^n \frac{1.3.5.7 \ldots (2n-1) }{2^n x^{n+1/2}} \end{split} \end{equation*}\]

\(\boldsymbol{y=10^{3-2x}}\)

Taking log both sides,

\[\begin{equation*} \begin{split} \log y &= (3-2x)\log 10 = 3 \log 10 - 2x \log 10\\ \end{split} \end{equation*}\]

Differentiating w.r.t \(x\),

\[\begin{equation*} \begin{split} \frac{1}{y}\frac{dy}{dx} &= 0 - 2 \log 10\\ y_1 &= -2\log 10 y = -2\log 10 \times {10}^{3-2x}\\ y_2 &= -2 \log 10 y_1 = 2 \log 10 \times 2\log 10 \times {10}^{3-2x}\\ y_3 &= -2 \log 10 y_2 = -2 \log 10 \times 2 \log 10 \times 2\log 10 \times {10}^{3-2x}\\ y_4 &= -2 \log 10 y_3 = 2 \log 10 \times 2 \log 10 \times 2 \log 10 \times 2\log 10 \times {10}^{3-2x}\\ & \vdots \\ y_n &= (-1)^n 2^n (\log 10)^n 10^{3-2x} \end{split} \end{equation*}\]

\(\boldsymbol{y = \frac{x}{a+bx}}\)

Breaking into partial fractions,

\[\begin{equation*} \begin{split} y &= \frac{1}{b} - \frac{a}{b(a+bx)}\\ \end{split} \end{equation*}\]

Differentiating w.r.t \(x\),

\[\begin{equation*} \begin{split} y_1 &= 0 - \frac{a}{b} \times (-1) (a+bx)^{-2} \times b = (-1)^2 \frac{a}{(a+bx)^2}\\ y_2 &= (-1)^2 a \times (-2)(a+bx)^{-3}\times b = (-1)^3 \frac{2ab}{(a+bx)^3}\\ y_3 &= (-1)^3 2ab \times (-3) (a+bx)^{-4} \times b = (-1)^4 \frac{6ab^2}{(a+bx)^4}\\ y_4 &= (-1)^4 6ab^2 \times (-4) (a + bx)^{-5} \times b = (-1)^5 \frac{24ab^3}{(a+bx)^5}\\ &= \vdots \\ y_n &= (-1)^{n+1}\frac{n!ab^{n-1}}{(a+bx)^{n+1}} \end{split} \end{equation*}\]

\(\boldsymbol{y = \frac{x^n}{x-1}}\)

Let \(x-1=t\), then \(x=t+1\).

The equation can be written as

\[\begin{equation*} \begin{split} y &= \frac{(t+1)^n}{t}\\ \end{split} \end{equation*}\]

Using binomial expansion,

\[\begin{equation*} \begin{split} y &= \frac{{}^{n}C_{0}t^n.1 + {}^{n}C_{1}t^{n-1}.1^2 + {}^{n}C_{2}t^{n-2}.1^3 + \ldots + {}^{n}C_{n}.1^n}{t}\\ y &= \frac{t^n + {}^{n}C_{1}t^{n-1}.1 + {}^{n}C_{2}t^{n-2}.1 + \ldots + 1}{t}\\ y &= t^{n-1} + {}^{n}C_{1}t^{n-2} + {}^{n}C_{2}t^{n-3} + \ldots + \frac{1}{t}\\ y &= (x-1)^{n-1} + {}^{n}C_{1}(x-1)^{n-2} + {}^{n}C_{2}(x-1)^{n-3} + \ldots + \frac{1}{(x-1)} \end{split} \end{equation*}\]

The \(\text{n}^{th}\) derivative is

\[\begin{equation*} \begin{split} y_n &= \frac{d^ny}{dx^n}\left[(x-1)^{n-1} + {}^{n}C_{1}(x-1)^{n-2} + {}^{n}C_{2}(x-1)^{n-3} + \ldots + \frac{1}{(x-1)}\right]\\ y_n &= 0 + 0 + 0 + \ldots + \frac{(-1)^n n!}{(x-1)^{n+1}}\\ &= \frac{(-1)^n n!}{(x-1)^{n+1}} \end{split} \end{equation*}\]

\(\boldsymbol{y = e^x \sin x \sin 2x}\)

\(\sin x \sin 2x\) can be written as

\[\begin{equation*} \begin{split} &= \frac{1}{2}[\cos (x-2x) - \cos 3x]\\ &= \frac{1}{2}[\cos(-x) - \cos 3x]\\ &= \frac{1}{2}[\cos x - \cos 3x] \end{split} \end{equation*}\]

Original equation can be written as thus

\[\begin{equation*} \begin{split} y &= e^x \sin x \sin 2x\\ &= e^x \times \frac{1}{2}[\cos x - \cos 3x]\\ &= \frac{1}{2}[e^x \cos x - e^x \cos 3x]\\ \end{split} \end{equation*}\]

The \(\text{n}^{th}\) derivative is

\[\begin{equation*} \begin{split} &= \frac{1}{2}\left[(1^2 + 1^2)^{n/2} e^x \cos (x + n\tan^{-1}(1/1)) - (1^2 + 3^2)^{n/2} e^x \cos (3x + n\tan^{-1}(3/1))\right]\\ &= \frac{1}{2}\left[2^{n/2} e^x \cos (x + \frac{n\pi}{4}) - 10^{n/2} e^x \cos (3x + n\tan^{-1}(3/1))\right]\\ &= \frac{1}{2}\left[2^{n/2} e^x \cos (x + \frac{n\pi}{4}) - 10^{n/2} e^x \cos (3x + n\tan^{-1}3)\right]\\ &= \frac{1}{2}e^x\left[2^{n/2} \cos (x + \frac{n\pi}{4}) - 10^{n/2} \cos (3x + n\tan^{-1}3)\right]\\ \end{split} \end{equation*}\]

\(\boldsymbol{y = e^{3x} \sin 4x}\)

This equation is of the form \(y=e^{ax}\sin bx\) where \(a=3\) and \(b=4\). So from the table above, we have,

\[\begin{equation*} \begin{split} y_n &= (a^2 + b^2)^{n/2} e^{ax} \sin \left( bx + n \tan^{-1} \frac{b}{a}\right)\\ y_n &= (3^2 + 4^2)^{n/2} e^{3x} \sin \left( 4x + n \tan^{-1} \frac{4}{3}\right)\\ &= (25)^{n/2} e^{3x} \sin \left( 4x + n \tan^{-1} \frac{4}{3}\right)\\ y_n &= 5^n e^{3x} \sin \left( 4x + n \tan^{-1} \frac{4}{3}\right)\\ \end{split} \end{equation*}\]

\(\boldsymbol{y = e^x \sin^2 x}\)

The equation can be written as \(y = e^x \sin x \sin x\).

We know from trigonometry,

\[\begin{equation*} \begin{split} \sin \alpha \sin \beta &= \frac{1}{2}[\cos (\alpha -\beta) - \cos(\alpha + \beta)] \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \sin x \sin x &= \frac{1}{2}[\cos (x -x) - \cos(x + x)]\\ &= \frac{1}{2}[1 - \cos2x]\\ \end{split} \end{equation*}\]

Thus the given equation can be written as

\[\begin{equation*} \begin{split} y &= e^x \sin x \sin x\\ y &= e^x \left(\frac{1}{2}[1 - \cos2x]\right)\\ y &= \frac{1}{2}[e^x - e^x \cos2x]\\ \end{split} \end{equation*}\]

And its \(\text{n}^{th}\) derivative is

\[\begin{equation*} \begin{split} y_n &= \frac{1}{2} \left[e^x - (1^2 + 2^2)^{n/2} e^{1x} \cos \left( 2x + n \tan^{-1} \frac{2}{1}\right)\right]\\ y_n &= \frac{1}{2} [e^x - 5^{n/2} e^x \cos( 2x + n \tan^{-1} 2)]\\ \end{split} \end{equation*}\]

5.1.2 Question 2

Find the \(\text{n}^{th}\) derivatives of the following functions:

\(\boldsymbol{y=\frac{1}{x^2 + 16}}\)

This equation can be written as \(y=\frac{1}{x^2 + 4^2}\) which is of the form \(y=\frac{1}{x^2 + a^2}\), so

\[\begin{equation*} \begin{split} y_n = \frac{(-1)^n n!}{4^{n+2}}\sin^{n+1}\theta \sin (n+1)\theta \end{split} \end{equation*}\]

where \(\theta = \tan^{-1}\left(\frac{4}{x}\right)\).

\(\boldsymbol{y = \frac{x}{x^2 + a^2}}\)

Breaking the equation into partial fractions for simplicity of calculating \(\text{n}^{th}\) derivative,

\[\begin{equation*} \begin{split} y &= \frac{x}{x^2 - (ia)^2}, \text{where } i = \sqrt{-1}\\ &= \frac{x}{(x+ia)(x-ia)}\\ &= \frac{1}{2}\left[\frac{1}{x+ia} + \frac{1}{x-ia}\right]\\ y_n &= \frac{1}{2}\left[D^n\left(\frac{1}{x+ia}\right) + D^n\left(\frac{1}{x-ia}\right)\right]\\ \end{split} \end{equation*}\]

The \(\text{n}^{th}\) derivative of \(\frac{1}{ax+b}\) is \(\frac{(-1)^n n! a^n}{(ax + b)^{n+1}}\).

\[\begin{equation*} \begin{split} y_n &= \frac{1}{2}\left[\frac{(-1)^n n!}{(x+ia)^{n+1}} + \frac{(-1)^n n!}{(x-ia)^{n+1}}\right]\\ &= \frac{(-1)^n n!}{2}\left[(x+ia)^{-(n+1)} + (x-ia)^{-(n+1)}\right]\\ \end{split} \end{equation*}\]

Lets put \(x = r\cos \theta\) and \(a = r\sin \theta\).

\[\begin{equation*} \begin{split} &= \frac{(-1)^n n!}{2}\left[(r\cos \theta+i r\sin \theta)^{-(n+1)} + (r\cos \theta-i r\sin \theta)^{-(n+1)}\right]\\ \end{split} \end{equation*}\]

Applying De-Moivre’s theorem,

\[\begin{equation*} \begin{split} y_n &= \frac{(-1)^n n!}{2}\left[r^{-(n+1)}\{ \cos(-(n+1)\theta) + i \sin(-(n+1)\theta)\} + \\ r^{-(n+1)}\{ \cos(-(n+1)\theta) - i \sin(-(n+1)\theta)\}\right]\\ y_n &= \frac{(-1)^n n!}{2r^{n+1}}[2 \cos (n+1)\theta]\\ y_n &= \frac{(-1)^n n!}{r^{n+1}}[\cos (n+1)\theta] \end{split} \end{equation*}\]

\(\cos (-\theta) = \cos \theta\)

Putting \(r = \frac{a}{\sin \theta}\),

\[\begin{equation*} \begin{split} y_n &= \frac{(-1)^n n!}{a^{n+1}} \sin^{n+1} \theta \cos (n+1)\theta \end{split} \end{equation*}\]

where \(\theta = \tan^{-1}\frac{a}{x}\).

\(\boldsymbol{y = \frac{1}{x^2 + x + 1}}\)

The \(x\) values of denominator from the formula \(\frac{-b \pm \sqrt{b^2 -4ac}}{2a}\) are \(\frac{-1 + \sqrt{3}i}{2}\) and \(\frac{-1 - \sqrt{3}i}{2}\).

So the equation can be resolved into partial fractions as follows,

\[\begin{equation*} \begin{split} y &= \frac{1}{\left[x - \left(\frac{-1 + \sqrt{3}i}{2}\right)\right]\left[x - \left(\frac{-1 - \sqrt{3}i}{2}\right)\right]}\\ &= \frac{4}{[(2x + 1) -\sqrt{3}i][(2x + 1) + \sqrt{3}i]}\\ y &= \frac{4}{2\sqrt{3}i}\left[\frac{1}{(2x + 1) -\sqrt{3}i} - \frac{1}{(2x + 1) + \sqrt{3}i}\right]\\ y &= \frac{2}{\sqrt{3}i}\left[\frac{1}{(2x + 1) -\sqrt{3}i} - \frac{1}{(2x + 1) + \sqrt{3}i}\right]\\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} y_n &= \frac{2}{\sqrt{3}i}\left[D^n\left(\frac{1}{(2x + 1) -\sqrt{3}i}\right) - D^n\left(\frac{1}{(2x + 1) + \sqrt{3}i}\right)\right]\\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} y_n &= \frac{2}{\sqrt{3}i}\left[\frac{(-1)^n n! 2^n}{(2x + 1 -\sqrt{3}i)^{n+1}} - \frac{(-1)^n n! 2^n}{(2x + 1 +\sqrt{3}i)^{n+1}}\right]\\ \end{split} \end{equation*}\]

The \(\text{n}^{th}\) derivative of \(\frac{1}{2x + a + b}\) is \(\frac{(-1)^n 2^n n!}{(2x + a +b)^{n+1}}\).

Lets put \(2x+1 = r \cos \theta\) and \(\sqrt{3} = r \sin \theta\), so that \(\tan \theta = \frac{\sqrt{3}}{2x + 1}\).

\[\begin{equation*} \begin{split} y_n &= \frac{(-1)^n 2^{n+1} n! }{\sqrt{3}i}\left[(r\cos \theta-i r\sin \theta)^{-(n+1)} - (r\cos \theta + i r\sin \theta)^{-(n+1)}\right]\\ \end{split} \end{equation*}\]

Applying De-Moivre’s theorem,

\[\begin{equation*} \begin{split} y_n &= \frac{(-1)^n 2^{n+1} n! }{\sqrt{3}i}\times r^{-(n+1)}\left[(\cos \theta-i \sin \theta)^{-(n+1)} - (\cos \theta + i \sin \theta)^{-(n+1)}\right]\\ &= \frac{(-1)^n 2^{n+1} n! }{\sqrt{3}i}\times r^{-(n+1)}\left[\{ \cos(-(n+1)\theta) - i \sin(-(n+1)\theta)\} - \\ \{ \cos(-(n+1)\theta) + i \sin(-(n+1)\theta)\}\right]\\ &= \frac{(-1)^n 2^{n+1} n! }{\sqrt{3}i}\times r^{-(n+1)}\left[\cos(n+1)\theta + i \sin(n+1)\theta - \\ \cos(n+1)\theta + i \sin(n+1)\theta\right]\\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} y_n &= &= \frac{(-1)^n 2^{n+1} n! r^{-(n+1)}}{\sqrt{3}i} [2i \sin (n+1)\theta]\\ y_n &= &= \frac{(-1)^n 2^{n+2} n!}{\sqrt{3}r^{(n+1)}} [ \sin (n+1)\theta]\\ \end{split} \end{equation*}\]

Substituting \(r = \frac{\sqrt{3}}{\sin \theta}\), we get

\[\begin{equation*} \begin{split} y_n &= \frac{(-1)^n 2^{n+2} n!}{\sqrt{3}\left(\frac{\sqrt{3}}{\sin \theta}\right)^{n+1}}[ \sin (n+1)\theta]\\ y_n &= \frac{(-1)^n 2^{n+2} n!}{(\sqrt{3})^{n+2}} \sin^{n+1}\theta \sin (n+1)\theta \end{split} \end{equation*}\]

where \(\theta = \tan^{-1}\left(\frac{\sqrt{3}}{2x+1}\right)\).

\(\boldsymbol{y = \tan^{-1}\frac{1+x}{1-x}}\)

Lets assume \(x=\tan \theta\).

Then equation becomes,

\[\begin{equation*} \begin{split} y &= \tan^{-1}\frac{\tan \pi/4 + \tan \theta}{1- \tan \pi/4 \tan \theta}\\ y &= \tan^{-1} \tan (\pi/4 + \theta)\\ y &= \frac{\pi}{4} + \theta \end{split} \end{equation*}\]

So \(n^{\text{th}}\) derivative is,

\[\begin{equation*} \begin{split} y_n &= D^n\left(\frac{\pi}{4}\right) + D^n(\tan^{-1}x)\\ y_n &= 0 + D^n(\tan^{-1}x)\\ y_n &= (-1)^{n-1} (n-1)! \sin^{n}\phi \sin n \phi \end{split} \end{equation*}\]

where \(\phi = \tan^{-1}\frac{1}{x}\).

The \(n^{\text{th}}\) derivative of \(\tan^{-1}x\) is \((-1)^{n-1} (n-1)! \sin^{n}\theta \sin n \theta\) where \(\theta = \tan^{-1}\frac{1}{x}\).

\(\boldsymbol{y = \frac{1}{(x^2+a^2)(x^2 + b^2)}}\)

Breaking the equation into derivable known partial fractions,

\[\begin{equation*} \begin{split} y &= \frac{1}{b^2-a^2}\left[\frac{1}{x^2 + a^2} - \frac{1}{x^2 + b^2}\right]\\ y_n &= \frac{1}{b^2-a^2}\left[D^n\left(\frac{1}{x^2 + a^2}\right) - D^n\left(\frac{1}{x^2 + b^2}\right)\right]\\ &= \frac{1}{b^2 -a^2}\left[\frac{(-1)^n n!}{a^{n+2}}\sin^{n+1}\theta \sin (n+1)\theta - \frac{(-1)^n n!}{b^{n+2}}\sin^{n+1}\phi \sin (n+1)\phi\right]\\ &= \frac{(-1)^n n!}{b^2 -a^2}\left[\frac{\sin^{n+1}\theta \sin (n+1)\theta}{a^{n+2}} - \frac{\sin^{n+1}\phi \sin (n+1)\phi}{b^{n+2}}\right]\\ y_n &= \frac{(-1)^n n!}{a^2 -b^2}\left[\frac{\sin^{n+1}\phi \sin (n+1)\phi}{b^{n+2}} - \frac{\sin^{n+1}\theta \sin (n+1)\theta}{a^{n+2}}\right]\\ \end{split} \end{equation*}\]

where \(\theta = \tan^{-1}\left(\frac{a}{x}\right)\) and \(\phi = \tan^{-1}\left(\frac{b}{x}\right)\)

\(\boldsymbol{y = \cot^{-1}\frac{x}{a}}\)

Differentiating w.r.t \(x\),

\[\begin{equation*} \begin{split} y_1 &= -\frac{a^2}{a^2 + x^2}\times \frac{1}{a}\\ y_1 &= (-1)\times \frac{a}{x^2 + a^2} \end{split} \end{equation*}\]

Differentiating the above equation \(n-1\) times gives \(n^{\text{th}}\) derivative. We can use general formula for \(y_n\) for expression \(\frac{1}{x^2 + a^2}\).

So in our case,

\[\begin{equation*} \begin{split} y_{1+n-1} &= (-1) D^{n-1}\left(\frac{a}{x^2 + a^2}\right)\\ y_{1+n-1} &= (-1) a \left[\frac{(-1)^{n-1} (n-1)!}{a^{n -1 +2}}\sin^{n -1 +1}\theta \sin (n -1 +1)\theta\right]\\ y_n &= (-1) a \left[\frac{(-1)^{n-1} (n-1)!}{a^{n + 1}}\sin^{n}\theta \sin (n)\theta\right]\\ &= \frac{(-1)^{n} (n-1)!}{a^n} \sin^{n}\theta \sin (n)\theta\\ \end{split} \end{equation*}\]

where \(\theta = \tan^{-1}\left(\frac{a}{x}\right)\)

\(\boldsymbol{y = \tan^{-1}\frac{\sqrt{1+x^2}-1}{x}}\)

Lets assume \(x= \tan \theta\).

So the equation becomes,

\[\begin{equation*} \begin{split} y &= \tan^{-1}\frac{\sqrt{1+\tan^2 \theta}-1}{\tan \theta}\\ &= \tan^{-1} \frac{\sec \theta -1}{\tan \theta}\\ &= \tan^{-1} \frac{1-\cos \theta}{\sin \theta}\\ &= \tan^{-1}\frac{2\sin^2 \theta/2}{2\sin \theta/2 \cos \theta/2}\\ &= \tan^{-1}\tan \frac{\theta}{2}\\ y &= \frac{\theta}{2}\\ y &= \frac{1}{2}\theta\\ y &= \frac{1}{2}\tan^{-1}x\\ \end{split} \end{equation*}\]

So \(n^{\text{th}}\) derivative is,

\[\begin{equation*} \begin{split} y_n &= \frac{1}{2}(-1)^{n-1} (n-1)! \sin^{n}\theta \sin n \theta \end{split} \end{equation*}\]

where \(\theta = \tan^{-1}\frac{1}{x}\).

5.1.3 Question 3

If \(y=\frac{1}{x^x}\), show that \(y_2(1) = 0\).

Taking log both sides,

\[\begin{equation*} \begin{split} \log y &= -x \log x \end{split} \end{equation*}\]

Differentiating w.r.t \(x\)

\[\begin{equation*} \begin{split} \frac{1}{y}\frac{dy}{dx} &= -x\frac{1}{x} - \log x\\ &= -1 - \log x \\ y_1 &= -y(1+\log x) \end{split} \end{equation*}\]

Again differentiating w.r.t \(x\),

\[\begin{equation*} \begin{split} \frac{dy_1}{dx} &= -\left[y_1(1+\log x) + \frac{y}{x}\right]\\ y_2 &= -\left[-y(1+\log x)(1 + \log x) + \frac{y}{x}\right]\\ y_2 &= y\left[(1+\log x)^2 - \frac{1}{x}\right]\\ y_2 &= \frac{1}{x^x}\left[(1+\log x)^2 - \frac{1}{x}\right]\\ \end{split} \end{equation*}\]

So,

\[\begin{equation*} \begin{split} y_2(1) &= \frac{1}{1}[(1+0)^2 - 1]\\ &= 1(1-1)\\ &= 0 \end{split} \end{equation*}\]

Hence proved.

5.1.4 Question 4

If \(\boldsymbol{ax^2 + 2hxy + by^2 =1}\), show that \(\boldsymbol{\frac{d^2y}{dx^2} = \frac{h^2 -ab}{(hx+by)^3}}\).

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\frac{2ax+2hy}{2hx+2by}\\ y_1 &= -\frac{ax+hy}{hx + by}\\ y_2 &= -\frac{[(hx+by)(a+hy_1) - (ax+hy)(h+by_1)]}{(hx+by)^2}\\ &= -\frac{[(hx+by)(a-\frac{h(ax+hy)}{hx+by})-(ax+hy)(h-\frac{b(ax+hy)}{hx+by})]}{(hx+by)^2}\\ &= -\frac{[(hx+by)\left(\frac{ahx+aby-ahx-h^2y}{hx+by}\right)-(ax+hy)\left(\frac{h^2x + bhy -abx - bhy}{hx+by}\right)]}{(hx+by)^2}\\ &= -\frac{[(hx+by)(aby -h^2y)-(ax+hy)(h^2x -abx)]}{(hx+by)^2\times (hx+by)}\\ &= \frac{x(ax+hy)(h^2 -ab) - y(ab-h^2)(hx+by)}{(hx+by)^3}\\ &= \frac{x(ax+hy)(h^2-ab) + y(h^2-ab)(hx+by)}{(hx+by)^3}\\ &= \frac{(h^2-ab)(ax^2 + hxy + hxy +by^2)}{(hx+by)^3}\\ &= \frac{(h^2-ab)(ax^2 + 2hxy + by^2)}{(hx+by)^3}\\ \frac{d^2y}{dx^2} &= \frac{h^2-ab}{(hx+by)^3}, \text{ as } ax^2 + 2hxy + by^2 = 1 \end{split} \end{equation*}\]

5.1.5 Question 5

If \(y=e^{-px}\cos qx\), prove that \(y_2 + 2py_1 + (p^2 +q^2)y = 0\).

Differentiating w.r.t \(x\),

\[\begin{equation*} \begin{split} y_1 &= -q e^{-px} \sin qx - p e^{-px} \cos qx\\ &= -q e^{-px} \sin qx - py \\ y_2 &= -q \frac{d}{dx}(e^{-px}\sin qx) -py_1 \\ y_2 &= -q [q e^{-px} \cos qx - p e^{-px} \sin qx] - py_1 \\ &= -q\left[qy + p \frac{y_1 + py}{q}\right] - py_1 \\ y_2 &= -\frac{q}{q}\left[q^2 y + py_1 + p^2 y\right] - py_1\\ y_2 + (p^2 + q^2)y + 2py_1 &= 0\\ \end{split} \end{equation*}\]

5.1.6 Question 6

If \(y=x^{2n}\), where \(n\) is a positive integer, show that \(\boldsymbol{y_n = 2^n[1.3.5\ldots (2n-3)(2n-1)]x^n}\).

\[\begin{equation*} \begin{split} y &= x^{2n}\\ y_1 &= 2n x^{2n-1}\\ y_2 &= 2n(2n-1)x^{2n-2}\\ y_3 &= 2n(2n-1)(2n-2)x^{2n-3}\\ & \vdots \\ y_n &= 2n (2n-1)(2n-2)(2n-3)(2n-4) \ldots (2n-n+1) x^{2n-n}\\ &= 2n (2n-1)(2n-2) (2n-3)(2n-4) \ldots (n+1) x^n\\ &= \frac{2n (2n-1)(2n-2)(2n-3)(2n-4)\ldots (n+1)n (n-1) (n-2) (n-3)\ldots 3.2.1}{n (n-1) (n-2) (n-3)\ldots 3.2.1}x^n\\ \end{split} \end{equation*}\]

Separating even and odd factors, we write

5.1.7 Question 7

If \(y = \sin mx + \cos mx\), prove that

\(\boldsymbol{y_2 + m^2y = 0}\)

\[\begin{equation*} \begin{split} y_1 &= m\cos mx - m\sin mx \\ y_2 &= -m^2 \sin mx - m^2 \cos mx\\ &= -m^2 (\sin mx + \cos mx)\\ &= -m^2 y \\ y_2 + m^2 y &= 0 \end{split} \end{equation*}\]

\(\boldsymbol{y_n = m^n \{1 + (-1)^n \sin 2mx\}^{1/2}}\)

We have, \(y = \sin mx + \cos mx\)

\[\begin{equation*} \begin{split} \frac{d^ny}{dx^n} &= D^n(\sin mx) + D^n(\cos mx)\\ y_n &= m^n \sin (mx + 0 + \frac{n\pi}{2}) + m^n \cos (mx + 0 + \frac{n\pi}{2})\\ &= m^n \left[\sin \left(mx + \frac{n\pi}{2}\right) + \cos \left(mx + \frac{n\pi}{2}\right)\right]\\ \end{split} \end{equation*}\]

Squaring and taking square root,

\[\begin{equation*} \begin{split} y_n &= m^n \left[\sqrt{\left\{\sin \left(mx + \frac{n\pi}{2}\right) + \cos \left(mx + \frac{n\pi}{2}\right)\right\}^2}\right]\\ &= m^n \left[\sqrt{1 + 2 \sin \left(mx + \frac{n\pi}{2}\right) \cos \left(mx + \frac{n\pi}{2}\right)}\right]\\ &= m^n \left[\sqrt{1 + \sin 2 \left(mx + \frac{n\pi}{2}\right)}\right]\\ &= m^n \left[\sqrt{1 + \sin (2mx + n\pi)}\right]\\ &= m^n \left[\sqrt{1 + \sin 2mx \cos n \pi + \cos 2mx \sin n \pi}\right]\\ &= m^n \{1 + (-1)^n \sin 2mx\}^{1/2} \end{split} \end{equation*}\]

because \(\cos n\pi = (-1)^n\) and \(\sin n \pi = 0\).

Thus,

\[\begin{equation*} \begin{split} y_n &= m^n \{1 + (-1)^n \sin 2mx\}^{1/2} \end{split} \end{equation*}\]