What is successive differentiation and what are the nth derivatives of e^(ax), a^(mx) and (ax+b)^m?

In one line

Successive differentiation means differentiating a function again and again; for standard functions we spot the pattern and write the \(n\)th derivative \(y_n\) as a formula in \(n\).

Explained simply

If you know the first few terms of a pattern, 2, 4, 8, 16, you can write the \(n\)th term as \(2^n\) without listing them all. Successive differentiation does the same with derivatives: compute \(y_1,y_2,y_3\), see the pattern, and write \(y_n\). This saves you from differentiating 50 times when someone asks for the 50th derivative.

Notation: \(y_1=\dfrac{dy}{dx}\), \(y_2=\dfrac{d^2y}{dx^2}\), …, \(y_n=\dfrac{d^ny}{dx^n}=D^ny\). The formulas below can be proved by mathematical induction.

Why it matters: higher derivatives appear in Taylor and Maclaurin series, in the analysis of vibrations and circuits, and in error estimates of numerical methods. In this module they also feed directly into Leibnitz's theorem, where you need the \(n\)th derivative of one factor in closed form.

How it works

1. \(y=e^{ax}\): \(y_1=ae^{ax}\), \(y_2=a^2e^{ax}\), so \(y_n=a^ne^{ax}\).
2. \(y=a^{mx}=e^{mx\log a}\): each derivative brings a factor \(m\log a\), so \(y_n=(m\log a)^na^{mx}\).
3. \(y=(ax+b)^m\): \(y_1=ma(ax+b)^{m-1}\), \(y_2=m(m-1)a^2(ax+b)^{m-2}\), so \(y_n=m(m-1)\cdots(m-n+1)a^n(ax+b)^{m-n}\).
4. If \(m\) is a positive integer: \(y_m=m!\,a^m\) (a constant) and \(y_n=0\) for \(n>m\).

Key formula

• \(D^n e^{ax}=a^ne^{ax}\)
• \(D^n a^{mx}=(m\log a)^na^{mx}\), \(a>0\)
• \(D^n(ax+b)^m=\dfrac{m!}{(m-n)!}a^n(ax+b)^{m-n}\) for integer \(m\ge n\); in general \(m(m-1)\cdots(m-n+1)a^n(ax+b)^{m-n}\)

\(a,b,m\) are constants, \(n\) a positive integer, log is natural log.

Quick example

• \(y=e^{3x}\): \(y_n=3^ne^{3x}\); \(y_{10}=3^{10}e^{3x}=59049e^{3x}\).
• \(y=2^{5x}\): \(y_n=(5\log2)^n2^{5x}\).
• \(y=(2x+1)^5\): \(y_3=5\cdot4\cdot3\cdot2^3(2x+1)^2=480(2x+1)^2\). Check by hand: \(y_1=10(2x+1)^4\), \(y_2=80(2x+1)^3\), \(y_3=480(2x+1)^2\). Correct.
• \(y=\sqrt{x}=x^{1/2}\): \(y_n=\dfrac12\left(\dfrac12-1\right)\cdots\left(\dfrac12-n+1\right)x^{\frac12-n}\).

Common mistakes

• Forgetting the factor \(a^n\) that comes from the inner function \(ax+b\) (chain rule).
• Writing \(D^na^{x}=a^x\); the factor \((\log a)^n\) is needed.
• Using \(\dfrac{m!}{(m-n)!}\) when \(m\) is a fraction or negative; then write the product form.

How MU asks it

The syllabus gives successive differentiation 3 hours, so expect short 3-mark parts like "Find the \(n\)th derivative of \(e^{2x}+3^{x}\)" or "Find \(y_n\) for \((3x-2)^{7}\)". These formulas are also building blocks in Leibnitz problems. For full marks write the standard result being used, then substitute.