Derivative as the slope of a tangent

Secant to tangent

Why is the derivative the slope of the tangent line?

Shrink the gap between two points on y = x² and watch the secant line turn into the tangent. The limit of its slope is the derivative.

Drag the ringed points on the board.

High school
University
Secant slope
Limit definition
Tangent line
Rate of change

x_0

1.00

h

h → 0

1.50

\frac{\Delta y}{h} \;\to\; f'(x_0)

$h$: 1.50
$\dfrac{\Delta y}{h}$: 3.500
$f'(x_0)$: 2.000
$\text{gap}$: 1.500

Secant vs tangentstill apart

Guided tour

1A wide secant
Start at $P=(1,1)$ with $h=1.5$ . The blue secant through $P$ and $Q$ rises $5.25$ over a run of $1.5$ , so its slope is $3.5$ , far from the tangent.
2Halve the gap
At $h=0.5$ the slope is $2.5$ . Every time $h$ shrinks, the secant swings closer to the dashed green tangent.
3Almost touching
At $h=0.05$ the slope reads $2.05$ . The pattern is exact: for $y=x^2$ at $x_0=1$ the secant slope is $2+h$ , so it approaches $2$ .
4Move the point
Slide to $x_0=-1$ . The tangent now falls: its slope is $-2$ . The derivative of $x^2$ is $2x$ , a new slope for every point.

Explanation

1
What a secant measures
A secant joins two points of a curve, $P=(x_0, f(x_0))$ and $Q=(x_0+h, f(x_0+h))$ . Its slope is the average rate of change over that interval:
$\frac{f(x_0+h)-f(x_0)}{h}$
It answers "how fast, on average, between these two points", not "how fast at $P$ ".
2
From secant to tangent
Let $Q$ slide toward $P$ by shrinking $h$ . The secants settle on one line, the tangent, and their slopes settle on one number. That number is the derivative:
$f'(x_0)=\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$
You cannot just set $h=0$ : the quotient becomes $\frac{0}{0}$ . The limit asks what the slopes approach, which is why the board never lets $h$ reach zero.
3
Worked example: y = x² at x = 1
Expand the difference quotient:
$\frac{(1+h)^2-1^2}{h}=\frac{2h+h^2}{h}=2+h$
As $h\to 0$ this goes to $2$ , so $f'(1)=2$ . The tangent passes through $(1,1)$ with slope $2$ :
$y-1=2(x-1)\quad\Longrightarrow\quad y=2x-1$
4
The derivative is a function
Repeat the computation at any $x_0$ : $\frac{(x_0+h)^2-x_0^2}{h}=2x_0+h\to 2x_0$ . So $(x^2)'=2x$ . Negative $x$ gives a falling tangent, $x=0$ gives a flat one, and the slope grows the further right you go.

Check yourself

Q01

For $f(x)=x^2$ at $x_0=1$ , what is the secant slope when $h=0.1$ ?

Q02

Why can we not simply substitute $h=0$ into the difference quotient?

Q03

What is the tangent line to $y=x^2$ at $x=-1$ ?

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