Power Rules and Linearity

Rules summarize a limit calculation

Repeatedly expanding a difference quotient is inefficient. Differentiation rules are results established from the definition; they let us calculate rates while preserving the underlying meaning. Start with constants, sums and powers before combining more complicated operations.

If $f$ and $g$ are differentiable and $c$ is constant, then $(cf+g)'=cf'+g'$ . A constant has derivative zero. For a real power on an interval where it is differentiable,

\frac{d}{dx}x^p=px^{p-1}.

For positive integer powers the rule follows by expanding $(x+h)^p$ : after subtraction and division, only $px^{p-1}$ survives the limit. Negative and fractional powers need attention to their domains.

Visual guide

VISUAL GUIDEA function and its slope function

f(x) = x³
f′(x) = 3x²

For f(x) = x³, the derivative is 3x². Negative x values still have positive slopes; the graph flattens at x = 0 where the derivative vanishes.

Worked example: rewrite first

Differentiate $f(x)=3x^4-2/x+5\sqrt{x}$ for $x>0$ . Rewrite as $3x^4-2x^{-1}+5x^{1/2}$ , then apply the rule term by term:

f'(x)=12x^3+2x^{-2}+\frac52x^{-1/2}.

The derivative is $12x^3+2/x^2+5/(2\sqrt{x})$ . Keeping $x>0$ avoids both the reciprocal singularity and the endpoint where the square-root derivative is unbounded. A symbolic expression for a derivative does not expand the original domain.

Worked example: rates of rates

For $s(t)=t^3-6t^2+9t$ meters, velocity is $v(t)=3t^2-12t+9$ m/s and acceleration is $a(t)=6t-12$ m/s². At $t=1$ , velocity is zero but acceleration is $-6$ m/s². A zero instantaneous velocity need not mean the object remains at rest.

The velocity factors as $3(t-1)(t-3)$ . For $t\ge0$ , the object moves in the positive direction before $1$ and after $3$ , and in the negative direction between those times. Differentiation supplies information; interpreting its signs supplies the physical conclusion.

What linearity does not say

Linearity applies to sums and constant multiples. It does not say $(fg)'=f'g'$ or $(f/g)'=f'/g'$ . Taking $f=g=x$ shows the first claim fails: $(x^2)'=2x$ , while $f'g'=1$ . Products, quotients and compositions require their own rules.

At an endpoint, a one-sided rate may exist even when a two-sided derivative is not defined. State which notion the problem needs instead of silently treating an endpoint as an interior point.

Practice

Differentiate $7x^5-3x+8$ .
Differentiate $x^{-3}$ and state its domain.
Find the second derivative of $x^4-2x^2$ .

Show worked solutions

$35x^4-3$ ; the constant contributes zero.
$-3x^{-4}=-3/x^4$ , for $x\ne0$ .
The first derivative is $4x^3-4x$ , and the second is $12x^2-4$ .