Differentiation Rules¶

Theorem¶

Let $I$ be an open interval, and let $f,g\colon I\to\mathbb{R}\text{.}$ At every point where the stated derivatives exist:

If $f$ is a constant function, then $f'(x)=0\text{.}$
If $f$ is the identity function $f(x)=x\text{,}$ then $f'(x)=1\text{.}$
(Power rule) For a positive integer $k\text{,}$ if $f(x)=x^k$ then $f'(x)=kx^{k-1}\text{.}$ To prove it from the definition, factor $x^k-x_0^k=(x-x_0)\left(x^{k-1}+x^{k-2}x_0+\cdots+x_0^{k-1}\right)\text{;}$ after dividing by $x-x_0\text{,}$ the second factor has $k$ terms, each tending to $x_0^{k-1}$ as $x\to x_0\text{.}$
(Sum and difference rule) If $f$ and $g$ are differentiable, then $f \pm g$ is differentiable, and $(f \pm g)'(x) = f'(x) \pm g'(x)\text{.}$
(Product rule) If $f$ and $g$ are differentiable, then $f g$ is differentiable, and $(fg)'(x) = f'(x)g(x) + f(x)g'(x)\text{.}$
If $f$ is differentiable and $c$ is a real number, then $cf$ is differentiable, and $(cf)'(x) = cf'(x)\text{.}$
If $g$ is differentiable and $g(x)\neq0\text{,}$ then $1/g$ is differentiable at $x\text{,}$ and $\displaystyle \left(\frac{1}{g}\right)'(x)=-\frac{g'(x)}{g(x)^2}\text{.}$
(Quotient rule) If $f$ and $g$ are differentiable and $g(x)\neq0\text{,}$ then $f/g$ is differentiable at $x\text{,}$ and

\left(\frac{f}{g}\right)'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}.

Interpretation¶

Differentiation rules compute derivatives without returning to the difference quotient each time: once the derivatives of a few basic functions are known, the derivative of any expression built from them by sums, products, quotients and constant multiples follows mechanically. Each rule is proved once from the definition of the derivative.

The sum and constant-multiple rules say that differentiation is linear: $(af+bg)'=af'+bg'$ for constants $a,b$ . The product rule is not of this form. Over a small step $h$ , the change in $fg$ consists of the change in $f$ times $g$ , plus $f$ times the change in $g$ , plus the product of the two changes. After division by $h$ the last term tends to $0$ , which leaves $f'g+fg'$ $f g^{'}$ . The quotient rule follows from the product rule and rule 7 by writing $f/g=f\cdot(1/g)$ $(1/ g)$ .

Repeated use of rules 1 to 6 shows that every polynomial is differentiable on $\mathbb{R}$ , with

(a_nx^n+\cdots+a_1x+a_0)' =na_nx^{n-1}+\cdots+a_1,

and the quotient rule extends this to rational functions wherever the denominator is nonzero. Together with the chain rule for compositions, the rules reduce the derivative of any elementary expression to the derivatives of its building blocks.

A common slip is to differentiate a product factor by factor. In general $(fg)'\neq f'g'$ $f^{'} g^{'}$ : for $f(x)=g(x)=x$ one has $f'(x)g'(x)=1$ $1$ , while $(x^2)'=2x$ $2 x$ . In the quotient rule the order in the numerator matters, since $f'g-fg'$ and $fg'-f'g$ differ by a sign.

Key Results¶

Differentiation rules preserve differentiability under algebraic operations.
The product and quotient rules account for interactions between factors.
Constant multiples factor out of derivatives.
Together with the chain rule, the standard formulas give

$\begin{gathered} (\ln x)'=\frac{1}{x}\;\;(x>0),\\ (b^x)'=b^x\ln b\;\;(b>0),\\ (x^p)'=p\,x^{p-1}\;\;(x>0,\ p\in\mathbb{R}). \end{gathered}$

The domain condition $x>0$ is essential for a general real exponent $p\text{;}$ for integer powers the power rule holds wherever the expression is defined.

Examples¶

$(x^2+3x)'=2x+3\text{.}$
$(x^2\sin x)'=2x\sin x+x^2\cos x\text{.}$

Prerequisites¶

Sources¶

Math I: Calculus · Lecture 6

Aliases: derivative rules, product rule, quotient rule, chain rule Keywords: sum rule, product rule, quotient rule, differentiate a product, differentiate a quotient, derivative of a composite function Symbols: element of, real numbers

Last updated: 22 October 2026 19:04