Find the n -th derivative of f(x) = 1/(a²-b²x²), a,b ≠0

Find the $n$-th derivative of $f(x) = \dfrac{1}{a^{2}-b^{2}x^{2}}$, $a,b\neq0$.

Step 1: Split into partial fractions

Factor the denominator and write

\[ \frac{1}{\bigl(a-bx\bigr)\bigl(a+bx\bigr)} = \frac{A}{a-bx}+\frac{B}{a+bx} \;\Longrightarrow\; 1 = A\bigl(a+bx\bigr)+B\bigl(a-bx\bigr). \]

Setting $x = a/b$ gives $1 = 2aA$; setting $x = -a/b$ gives $1 = 2aB$. Hence $A = B = \dfrac{1}{2a}$ and

\[ f(x) = \frac{1}{2a}\left(\frac{1}{a-bx}+\frac{1}{a+bx}\right). \tag{1} \]

Step 2: The derivative of a simple pole

Repeated use of the power and chain rules gives, for constants $c\neq0$ and $d$, \[\frac{d^{n}}{dx^{n}}\left(\frac{1}{cx+d}\right) = \frac{\bigl(-1\bigr)^{n}\,n!\,c^{n}}{\bigl(cx+d\bigr)^{n+1}}. \tag{2}\]

Step 3: Apply (2) to each piece

For the first fraction $c = -b$, and the two sign factors combine to $\bigl(-1\bigr)^{2n} = 1$: \[\left(\frac{1}{a-bx}\right)^{(n)} = \frac{\bigl(-1\bigr)^{n}n!\bigl(-b\bigr)^{n}}{\bigl(a-bx\bigr)^{n+1}} = \frac{n!\,b^{n}}{\bigl(a-bx\bigr)^{n+1}}.\]

For the second $c = +b$, and the sign survives: \[\left(\frac{1}{a+bx}\right)^{(n)} = \frac{\bigl(-1\bigr)^{n}n!\,b^{n}}{\bigl(a+bx\bigr)^{n+1}}.\]

Step 4: Combine

\[ f^{(n)}(x) = \frac{n!\,b^{n}}{2a}\left[\frac{1}{\bigl(a-bx\bigr)^{n+1}}+\frac{\bigl(-1\bigr)^{n}}{\bigl(a+bx\bigr)^{n+1}}\right]. \]

Step 5: Verification

\[n = 0:\ \frac{1}{2a}\left[\frac{1}{a-bx}+\frac{1}{a+bx}\right] = \frac{1}{2a}\cdot\frac{2a}{a^{2}-b^{2}x^{2}} = f(x) \;\;\checkmark\]

\[n = 1:\ \frac{b}{2a}\left[\frac{1}{\bigl(a-bx\bigr)^{2}}-\frac{1}{\bigl(a+bx\bigr)^{2}}\right] = \frac{b}{2a}\cdot\frac{4abx}{\bigl(a^{2}-b^{2}x^{2}\bigr)^{2}} = \frac{2b^{2}x}{\bigl(a^{2}-b^{2}x^{2}\bigr)^{2}}\]

\[\text{and direct differentiation of } f \text{ gives the same } \;\;\checkmark\]