The Derivative from First Principles

From a time interval to an instant

If a position changes from $s(a)$ to $s(a+h)$ over a nonzero time interval $h$ , its average velocity is the displacement divided by elapsed time. The derivative is the limiting average rate as that interval shrinks:

f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}h.

A finite two-sided limit is required at an interior point. A derivative has units of output per input: meters per second for position against time, or amperes per volt for a current-voltage relationship.

The quotient is the slope of a secant through two graph points. Its limit, when it exists, is the tangent slope at the base point. A tangent is a local linear model, not necessarily a line that touches the graph only once.

Worked example: derive rather than memorize

For $f(x)=x^2$ ,

\frac{(a+h)^2-a^2}{h}=\frac{2ah+h^2}{h}=2a+h\quad(h\ne0).

Taking the limit gives $f'(a)=2a$ . At $a=3$ , the point is $(3,9)$ and the tangent slope is $6$ , so the tangent line is $y-9=6(x-3)$ . This calculation distinguishes the function value $9$ from its local rate $6$ .

If $s(t)=t^2$ meters for time in seconds, $s'(3)=6$ m/s. It is not a distance and does not mean the position increases by exactly six meters over every following second. The rate changes with time.

Worked example: a corner

For $f(x)=|x|$ at zero, the quotient is $|h|/h$ . It equals $1$ for positive $h$ and $-1$ for negative $h$ , so the derivative does not exist. The function is nevertheless continuous. An infinite limiting slope, as with $x^{1/3}$ at zero, is also not a finite derivative.

Differentiability implies continuity. Indeed, $f(a+h)-f(a)$ is $h$ times the difference quotient; if the quotient has a finite limit, this product tends to zero. This gives a useful preliminary check: a discontinuous function cannot be differentiable at that point.

Read derivative notation carefully

$f'(a)$ is a number at one input; $f'(x)$ describes the derivative function. The notation $dy/dx$ records which variables supply the rate. A second derivative measures how the first rate changes and has units of output per input squared.

Practice

Use the definition to differentiate $f(x)=4x+1$ .
Find the tangent to $x^2$ at $x=-1$ .
Does a finite difference quotient with $h=0.01$ always equal the derivative?

Show worked solutions

The numerator becomes $4h$ , so the quotient and its limit are $4$ .
The point is $(-1,1)$ and slope is $-2$ . Thus $y-1=-2(x+1)$ , or $y=-2x-1$ .
No. It is an average rate over a finite interval. For $x^2$ its error from the derivative is exactly $h$ .

Explore

From secant to tangent

Try this. Hold x fixed and shrink h: the orange secant approaches the green tangent. Then change x to compare local rates.

Base point x0.5Secant step h0.5

f(x) = x². Secant slope = ((x+h)² − x²)/h = 2x+h = 1.5. Tangent slope 2x = 1. Difference = h = 0.5. Orange: secant; green: tangent.