What Is the nth Derivative of (ax + b)^(-m)?

For constants $a$, $b$, and $m$, the expression $(ax+b)^{-m}$ combines a power rule with the chain rule. Each differentiation contributes another factor of $a$, changes the sign, and increases the magnitude of the exponent in the denominator.

For a positive integer $m$ and a nonnegative integer $n$, the formula is

$$\frac{d^n}{dx^n}(ax+b)^{-m} = (-1)^n a^n m(m+1)\cdots(m+n-1)(ax+b)^{-m-n}.$$

The product has $n$ factors. When $n=0$, it is an empty product, defined to be $1$, so the formula gives the original function.

Why the formula works

Start with the ordinary power rule and chain rule:

$$\frac{d}{dx}(ax+b)^{-m} = -m(ax+b)^{-m-1}\cdot a = -am(ax+b)^{-m-1}.$$

Differentiate once more. The exponent is now $-m-1$, so another application of the chain rule gives

$$\frac{d^2}{dx^2}(ax+b)^{-m} = a^2m(m+1)(ax+b)^{-m-2}.$$

A third derivative introduces the next factor, $m+2$, and another negative sign:

$$\frac{d^3}{dx^3}(ax+b)^{-m} = -a^3m(m+1)(m+2)(ax+b)^{-m-3}.$$

After $n$ derivatives, the sign is $(-1)^n$, the chain rule has supplied $a^n$, and the factors from differentiating the power are $m$ through $m+n-1$. This gives the stated result.

You can also justify the pattern by induction: differentiating the formula for the $n$th derivative multiplies it by $-(m+n)a$ and decreases the exponent by one. That produces exactly the formula with $n$ replaced by $n+1$.

Worked example

Find the third derivative of

$$f(x)=(3x-2)^{-4}.$$

Here $a=3$, $m=4$, and $n=3$. Substitute into the formula:

$$\begin{aligned} f^{(3)}(x) &= (-1)^3 3^3\cdot 4\cdot 5\cdot 6\,(3x-2)^{-4-3}\\ &= -27\cdot120\,(3x-2)^{-7}\\ &= -3240(3x-2)^{-7}. \end{aligned}$$

Thus, equivalently,

$$f^{(3)}(x)=\frac{-3240}{(3x-2)^7}.$$

The domain excludes $x=\frac{2}{3}$, where the original function is undefined. For $x>\frac{2}{3}$, the first derivative is negative, the second positive, and the third negative; the coefficient signs alternate with the order of differentiation.

A compact notation for the coefficient

The rising factorial is defined by

$$(m)^{\overline{n}}=m(m+1)\cdots(m+n-1),$$

with $(m)^{\overline{0}}=1$. Using it, the formula becomes

$$\frac{d^n}{dx^n}(ax+b)^{-m} = (-1)^n a^n(m)^{\overline{n}}(ax+b)^{-m-n}.$$

For positive real $m$, the product can also be written with the gamma function as $\Gamma(m+n)/\Gamma(m)$. The product form is usually simplest when $n$ is a whole number.

Common mistakes

• Forgetting the chain-rule factor. Differentiating with respect to $x$ contributes a factor of $a$ each time, so the $n$th derivative has $a^n$, not just $a$.
• Using the wrong product. The coefficient is $m(m+1)\cdots(m+n-1)$, not $m^n$ and not the descending product $m(m-1)\cdots$.
• Missing the alternating coefficient sign. The coefficient contains $(-1)^n$: it is negative for odd $n$ and positive for even $n$, though the sign of the full derivative can also depend on $ax+b$.
• Miscounting the exponent. It becomes $-m-n$ after $n$ derivatives.
• Ignoring where the function is defined. If $a\ne0$ and $m$ is a positive integer, exclude the point where $ax+b=0$. If $m$ is a general real number, the real-valued power may require $ax+b>0$, depending on $m$; use a domain on which the original function is defined.

Special cases

If $a=0$ and $b\ne0$, the expression is constant in $x$, so every derivative of positive order is zero. The general formula also gives zero for $n\ge1$ because it contains $a^n$. If $a=0$ and $b=0$, the expression is undefined for positive $m$.

If $m$ is a positive integer, the formula applies directly wherever $ax+b\ne0$. For other real values of $m$, the same differentiation pattern holds on any interval where $(ax+b)^{-m}$ is a well-defined real-valued function. The algebraic formula does not remove the need to check that domain.