In one line
Break the rational function into simple partial fractions like \(\dfrac{A}{x-a}\), then apply \(D^n\dfrac1{ax+b}=\dfrac{(-1)^nn!a^n}{(ax+b)^{n+1}}\) to each piece and add.
Explained simply
Differentiating \(\dfrac{x}{(x-1)(x-2)}\) \(n\) times directly is a nightmare of quotient rules. But it can be split into \(\dfrac{-1}{x-1}+\dfrac{2}{x-2}\), and each simple fraction has a known \(n\)th derivative. It is like breaking a big task into small ones you already know how to do.
Technically, a proper rational function \(\dfrac{P(x)}{Q(x)}\) with \(Q\) having distinct linear factors can be written as \(\sum\dfrac{A_k}{x-a_k}\). If the degree of \(P\) is not less than that of \(Q\), divide first: the polynomial part disappears after enough differentiations.
The cover-up rule is the fastest way to find the constants: to find the constant over \((x-a)\), cover that factor in the original fraction and put \(x=a\) in what remains. Always check by recombining the fractions, because one wrong constant spoils the entire answer.
How it works
- If needed, divide to make the fraction proper.
- Factorise the denominator.
- Write partial fractions and find constants (cover-up rule: put \(x=a_k\)).
- Apply the standard result to each fraction.
- For a factor like \(x^2+a^2\), split into \(\dfrac{1}{(x+ia)(x-ia)}\), or use \(\dfrac{1}{x^2+a^2}\) with the substitution \(x=a\cot\theta\).
Key formula
\(D^n\dfrac{A}{x-a}=\dfrac{(-1)^nn!\,A}{(x-a)^{n+1}}\); for a repeated factor, \(D^n\dfrac1{(x-a)^2}=\dfrac{(-1)^n(n+1)!}{(x-a)^{n+2}}\).
Quick example
\(y=\dfrac{x}{(x-1)(x-2)}\).
Cover-up: at \(x=1\), \(A=\dfrac{1}{1-2}=-1\); at \(x=2\), \(B=\dfrac{2}{2-1}=2\). So \(y=\dfrac{-1}{x-1}+\dfrac{2}{x-2}\).
\(y_n=(-1)^nn!\left[\dfrac{-1}{(x-1)^{n+1}}+\dfrac{2}{(x-2)^{n+1}}\right]\).
Check \(n=0\) gives back \(y\): \(\dfrac{-1}{x-1}+\dfrac2{x-2}=\dfrac{-(x-2)+2(x-1)}{(x-1)(x-2)}=\dfrac{x}{(x-1)(x-2)}\). Correct.
Common mistakes
- Wrong partial fraction constants; always recombine to check.
- Applying the formula to an improper fraction without dividing first.
- For \((x-a)^2\) in the denominator, using \(n!\) instead of \((n+1)!\).
How MU asks it
Very common 5/6-mark parts: "Find the \(n\)th derivative of \(\dfrac{x^2}{(x+2)(2x+3)}\)" or "\(\dfrac{1}{x^2-a^2}\)" (answer \(\dfrac{(-1)^nn!}{2a}\left[\dfrac1{(x-a)^{n+1}}-\dfrac1{(x+a)^{n+1}}\right]\)). Show the division (if any), the partial fractions with constants found, the standard result, and the final combined answer.