In one line
\(D^n\dfrac1{ax+b}=\dfrac{(-1)^nn!\,a^n}{(ax+b)^{n+1}}\) and \(D^n\log(ax+b)=\dfrac{(-1)^{n-1}(n-1)!\,a^n}{(ax+b)^n}\); signs alternate and factorials build up with each derivative.
Explained simply
Each time you differentiate \(\dfrac1{ax+b}\), the power in the denominator goes up by one, a new number multiplies in front, and the sign flips. Watching three steps is enough to see the pattern: \(-1\), then \(+2\), then \(-6\), which is \((-1)^n n!\).
The log result follows because \(\dfrac{d}{dx}\log(ax+b)=\dfrac{a}{ax+b}\). So the \(n\)th derivative of the log is \(a\) times the \((n-1)\)th derivative of \(\dfrac1{ax+b}\). These two formulas, together with partial fractions, let you find \(y_n\) for any rational function.
A quick way to remember them: the reciprocal formula has \(n!\) and power \(n+1\); the log formula is "one step behind", so it has \((n-1)!\) and power \(n\). The sign factor \((-1)^n\) or \((-1)^{n-1}\) simply records how many times a minus sign has appeared.
How it works
- \(y=(ax+b)^{-1}\Rightarrow y_1=-a(ax+b)^{-2}\), \(y_2=2a^2(ax+b)^{-3}\), \(y_3=-6a^3(ax+b)^{-4}\).
- Pattern: \(y_n=(-1)^nn!\,a^n(ax+b)^{-(n+1)}\).
- For \(y=\log(ax+b)\): \(y_1=a(ax+b)^{-1}\). Apply step 2 with \(n-1\) in place of \(n\): \(y_n=a\cdot(-1)^{n-1}(n-1)!\,a^{n-1}(ax+b)^{-n}\).
- Simplify: \(y_n=\dfrac{(-1)^{n-1}(n-1)!\,a^n}{(ax+b)^n}\).
Key formula
- \(D^n\dfrac{1}{ax+b}=\dfrac{(-1)^nn!\,a^n}{(ax+b)^{n+1}}\)
- \(D^n\log(ax+b)=\dfrac{(-1)^{n-1}(n-1)!\,a^n}{(ax+b)^n}\)
\(n!\) is \(n\) factorial; log is the natural logarithm; valid where \(ax+b\ne0\) (and \(ax+b>0\) for the log).
Quick example
- \(y=\dfrac1{2x+3}\): \(y_n=\dfrac{(-1)^nn!\,2^n}{(2x+3)^{n+1}}\). For \(n=2\): \(\dfrac{2\cdot4}{(2x+3)^3}=\dfrac{8}{(2x+3)^3}\). Direct check: \(y_1=-2(2x+3)^{-2}\), \(y_2=8(2x+3)^{-3}\). Correct.
- \(y=\log(1-x)\) (\(a=-1\)): \(y_n=\dfrac{(-1)^{n-1}(n-1)!(-1)^n}{(1-x)^n}=-\dfrac{(n-1)!}{(1-x)^n}\).
- \(y=\log(x^2-a^2)=\log(x-a)+\log(x+a)\): \(y_n=(-1)^{n-1}(n-1)!\left[\dfrac1{(x-a)^n}+\dfrac1{(x+a)^n}\right]\).
Common mistakes
- Using \((n+1)\) as the power in the log formula; it is \(n\) for log and \(n+1\) for the reciprocal.
- Forgetting \(a^n\), especially when \(a=-1\), which changes the sign.
- Not splitting \(\log\) of a product into a sum before differentiating.
How MU asks it
3-mark Q.1 parts: "Find the \(n\)th derivative of \(\log(ax+b)\)" or "of \(\log\left[(2x+3)^3(x-1)\right]\)" (split using log rules first). They are also the core of partial-fraction problems worth 5/6 marks. Quote the standard result, show how \(a\) and \(b\) are identified, and simplify.