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Grade 12 Calculus · Lesson 2.190 min

The Derivative & the Tangent Line Problem

Start with a familiar slope, let a secant line approach a tangent line, build the derivative from a limit, and use that same idea to measure instantaneous motion.

Build the derivative from slope and average rate of change, move from secant to tangent lines, use the limit definition, and interpret instantaneous velocity.

Slope → rateSecant → tangentDerivative at a pointFirst principlesInstantaneous velocity

Slope → Average Rate of Change

The derivative begins with an idea you already know: slope measures a rate of change. Between two points on a graph, that rate is an average rate of change.

m=\frac{y_2-y_1}{x_2-x_1}=\frac{f(x_2)-f(x_1)}{x_2-x_1}=\frac{\Delta y}{\Delta x}

Slope→

Average rate→

Secant→

Tangent→

Derivative

Worked example · Average rate

For $f(x)=x^2+1$ , find the average rate of change from $x=1$ to $x=4$ .

Your turn

Which expression gives the average rate of change of

f

from

x_1

x_2

, where

x_1\ne x_2

f(x_2)-f(x_1)

\frac{f(x_2)-f(x_1)}{x_2-x_1}

\frac{x_2-x_1}{f(x_2)-f(x_1)}

Secant Line → Tangent Line

Definition

Secant line

A secant line passes through two points on a curve. Its slope is an average rate of change.

Explanation

Keep one point fixed and move the second point toward it. Then $\Delta x\to0$ . The secant slopes approach one limiting value: the slope of the tangent line.

Interactive lab

From secant to tangent

Predict the limiting slope, then move Q toward P on

f(x)=\frac12x^2+0.65

. P stays at x = 1.

h = 2.00 · Secant: 2.000

Horizontal change h

As h approaches 0, which slope does the secant approach?

Worked example

Numerical bridge · secant to tangent

For $f(x)=x^2$ at $P=(1,1)$ , let $Q=(1+h,(1+h)^2)$ . Predict what happens to the secant slope as $h$ gets smaller.

m_{sec}=\frac{(1+h)^2-1}{h}=2+h

$h$	1	0.5	0.1	0.01
$m_{sec}$	3	2.5	2.1	2.01

m_{\text{tangent}}=\lim_{\Delta x\to0}m_{\text{secant}}

This is the geometric heart of the derivative: a tangent slope is not found from two distinct fixed points; it is obtained as the limit of secant slopes.

Your turn

As the second point on a secant approaches the point of tangency, what happens to

\Delta x

\Delta x=1

\Delta x\to\infty

\Delta x\to0

Derivative at a Point & the Tangent Line Problem

Writing the horizontal change as $h$ turns the secant-slope idea into the derivative. At a specific input $x=a$ , the derivative $f'(a)$ is the tangent slope at $(a,f(a))$ .

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

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

Explanation

The tangent-line chain

f'(a)

→tangent slope→

m=f'(a)

→

y-f(a)=f'(a)(x-a)

An equivalent definition

Instead of describing the second input as $x+h$ , call it $w$ . As $w\to x$ , the two inputs come together. The quotient is the same secant slope written with different symbols.

f'(x)=\lim_{w\to x}\frac{f(w)-f(x)}{w-x}

f'(x)

prime notation

y'

prime notation for y

\frac{dy}{dx}

Leibniz notation

Common mistake. Do not put

h=0

into the difference quotient before simplifying. The derivative is a limit as

h

approaches zero.

Explanation

Derivative at a point → tangent line

At $x=a$ , $f'(a)$ is the slope of the tangent line at $(a,f(a))$ .

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

y-f(a)=f'(a)(x-a)

Worked example · Tangent line

For $g(x)=x^2+1$ , find the tangent line at $x=2$ .

Worked example

Equivalent limit form · worked from start to finish

For $f(x)=x^2$ , use the $w\to x$ form.

f'(x)=\lim_{w\to x}\frac{w^2-x^2}{w-x}

Your turn

For

f(x)=x^2

, what should you write first when using the

h

-definition?

f(x+h)=(x+h)^2

f(x+h)=x^2+h

f(x+h)=2x+h

Mr. Azzam’s Routine · Derivative from Definition

STEP 1

Write the definition

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

STEP 2

Find f(x+h)

Replace every

x

(x+h)

STEP 3

Substitute carefully

Build

f(x+h)-f(x)

using brackets.

STEP 4

Simplify vertically

Expand or factor until a factor

h

cancels.

STEP 5

Apply the limit

Only after simplifying, let

h\to0

STEP 6

Interpret

State what the result means: tangent slope or instantaneous rate, with units when appropriate.

Your turn

For

f(x)=x^2

, the difference quotient is

\frac{2xh+h^2}{h}

for

h\ne0

. What is the correct next step?

Set

h=0

immediatelyCancel only one

h

and keep

2xh+h

Conclude the derivative is undefinedFactor

h

, cancel it, then take

h\to0

Worked Examples · Build the Derivative Yourself

These examples move from linear to quadratic to cubic. Try each checkpoint in order, or reveal the complete worked solution when you need support. After the algebra, connect the result back to tangent slope.

Worked example

Example 1 · Linear

f(x)=3-5x

Step 1Step 2Step 3Step 4Step 5

First find

f(x+h)

Worked example

Example 2 · Quadratic

f(x)=x^2+x-3

Step 1Step 2Step 3Step 4Step 5

Find

f(x+h)

Worked example

Example 3 · Cubic

f(x)=x^3-12x

Step 1Step 2Step 3Step 4Step 5

Find

f(x+h)

Key takeaway

Geometric interpretation

A derivative formula is not just an algebraic result. Once you know $f'(x)$ , substituting a particular input gives the slope of the tangent line there. If a question also asks for the tangent line, pair that slope with the point $(a,f(a))$ .

Your turn

For

f(x)=x^2+x-3

, which is the tangent line at

x=2

y=5x

y-5=3(x-2)

y-3=5(x-2)

y=3

Rates of Change & Instantaneous Velocity

Worked example

AlHamra Tower problem

A ball is dropped from the upper observation deck, 200 m above the ground. Find its velocity after 5 seconds.

Explanation

Average velocity

Two times are used, so this is a secant slope.

v_{avg}=\frac{s(t+h)-s(t)}{h}

Explanation

Instantaneous velocity

Let the time interval shrink to zero: the secant slope becomes a tangent slope.

v(t)=s'(t)=\lim_{h\to0}\frac{s(t+h)-s(t)}h

Worked example

Complete application · AlHamra Tower

Using $s(t)=200-4.9t^2$ , find the instantaneous velocity after $5$ seconds from the definition.

Your setup: Which limit represents $v(5)$ ?

$v(5)=\displaystyle\lim_{h\to0}\frac{s(5+h)-s(5)}h$

Key takeaway

Interpret the result, not just the number.

Velocity carries units of distance per time. A negative velocity means the position is decreasing in the chosen positive direction; in the tower model, that means the ball is moving downward.

Your turn

s(t)

is position, what does

s'(5)

represent?

Position at

t=5

Instantaneous velocity at

t=5

Average velocity from

0

5

Exam Strategy & Common Mistakes

Key takeawayWhen the question says “using the definition”

Show the limit definition → calculate $f(x+h)$ → form the numerator with brackets → simplify until $h$ cancels → apply the limit → interpret the result if the context asks for it.

Key takeawayWhen the question asks for a tangent line

Find $f'(a)$ for the slope, find $f(a)$ for the point, then use $y-f(a)=f'(a)(x-a)$ .

Common mistake. Do not write

f(x+h)=f(x)+f(h)

. Substitute

(x+h)

into the original function.

Common mistake. Do not cancel terms across addition. Simplify until

h

is a factor of the whole numerator.

Common mistake. Do not confuse average velocity with instantaneous velocity. Average velocity uses a nonzero time interval; instantaneous velocity is the limit as that interval approaches zero.

Your turn

The definition gives

f'(x)=-2/x^3

for

f(x)=1/x^2

. Where does this derivative apply?

All real

x

Only

x>0

Only

x<0

All real

x\ne0

Slope → Average Rate of Change

Your turn

Your turn · student questions

Secant Line → Tangent Line

Numerical bridge · secant to tangent

Your turn

Your turn · student questions

Derivative at a Point & the Tangent Line Problem

An equivalent definition

Derivative at a point → tangent line

Equivalent limit form · worked from start to finish

Your turn

Your turn · student questions

Mr. Azzam’s Routine · Derivative from Definition

Write the definition

Find f(x+h)

Substitute carefully

Simplify vertically

Apply the limit

Interpret

Your turn

Your turn · student questions

Worked Examples · Build the Derivative Yourself

Step 1

Step 2

Step 3

Step 4

Step 5

Step 1

Step 2

Step 3

Step 4

Step 5

Step 1

Step 2

Step 3

Step 4

Step 5

Your turn

Your turn · student questions

Rates of Change & Instantaneous Velocity

Your turn

Your turn · student questions

Exam Strategy & Common Mistakes

Your turn

Your turn · student questions