For a straight line, the slope measures its steepness and direction. Given two points:

the slope is:

We can describe this as:

A straight line has the same slope everywhere. A curve can have different slopes at different points. For example, the graph of:

falls on its left side, becomes horizontal at its lowest point, and rises on its right side. To describe a curve’s slope at a particular point, we use a tangent line . The central idea of this lesson is:

A secant line is a straight line through two distinct points on a curve. Suppose the points are:

and:

The secant slope is:

This is the function’s average rate of change between the two inputs.

A tangent line represents the curve’s local direction at a point. For a differentiable function, if we zoom in sufficiently near that point, the curve becomes approximately straight and follows its tangent line.

From a secant to a tangent

Keep the point:

fixed. Choose a nearby point:

The secant slope is:

As \(h\) approaches zero, \(Q\) approaches \(P\) . If the secant slopes approach a finite value, that value is the tangent slope:

This is the definition of the derivative at \(x=a\) :

Important: We approach \(h=0\) . We do not substitute \(h=0\) into a fraction whose denominator is \(h\) .

A tangent can cross a curve

A tangent is not defined as a line that “touches the curve only once.” For example, the tangent to:

at the origin is:

The curve crosses this tangent at the origin. What matters is the tangent’s local direction, not the number of intersections.

Consider:

We will find the tangent slope at:

The fixed point is:

A nearby point is:

The secant slope is:

Expand:

Therefore:

For \(h\neq0\) , simplify:

As \(h\) approaches zero:

Therefore:

The notation:

represents the derivative at one particular input. The notation:

represents a function that gives the tangent slope at different inputs. For example:

has derivative:

Therefore:

These are slopes, not heights.

Function value versus derivative value

At \(x=3\) :

whereas:

This means:

• The point on the original curve is \((3,9)\).
• The tangent slope there is \(6\).
• The point on the derivative graph is \((3,6)\).

The function and its derivative describe different quantities.

Simple derivative rules used in this lesson

For a constant \(c\) :

For a linear function:

For a positive integer power:

Differentiate polynomial terms separately. For example:

gives:

6.1 Positive derivative

If:

the tangent line at \(x=a\) rises from left to right. At that point, the function has a positive instantaneous rate of change.

6.2 Negative derivative

If:

the tangent line falls from left to right. The function has a negative instantaneous rate of change there.

6.3 Zero derivative

If:

the tangent line is horizontal. The point is called a stationary point . A zero derivative does not, by itself, tell us whether the point is a maximum, minimum, or neither.

Comparing steepness

On axes with the same scale, a larger absolute slope means a steeper tangent. For example:

A tangent with slope \(-5\) is steeper than one with slope \(2\) , although it slopes downward.

To find the tangent slope at \(x=a\) :

1. Differentiate the function.
2. Substitute \(x=a\) into the derivative.

A straight line through \((x_0,y_0)\) with slope \(m\) has equation:

At the point:

the tangent slope is:

Therefore, the tangent equation is:

You need two pieces of information:

• The point on the curve.
• The derivative value at that point.

A horizontal line has slope zero. Therefore, horizontal tangents occur where:

To find stationary points:

1. Differentiate.
2. Solve \(f'(x)=0\).
3. Substitute each solution into the original function.
4. Report the coordinates.

A stationary point need not be a turning point

Consider:

Its derivative is:

At \(x=0\) :

The tangent at the origin is horizontal:

However, the function continues increasing through the origin. It does not change from increasing to decreasing or vice versa. Therefore, the origin is stationary but is neither a local maximum nor a local minimum.

The derivative graph records the slopes of the original graph.

Reading the sign of a derivative graph

• Where \(f'(x)\) is above the \(x\)-axis, the original graph has positive tangent slopes.
• Where \(f'(x)\) is below the \(x\)-axis, the original graph has negative tangent slopes.
• Where \(f'(x)=0\), the original graph has a horizontal tangent.

If \(f'(x)>0\) throughout an interval, \(f\) is strictly increasing there. If \(f'(x)<0\) throughout an interval, \(f\) is strictly decreasing there.

What a derivative graph does not tell you directly

Knowing tangent slopes does not determine the original graph’s vertical position. For example:

and:

both have derivative:

Their slopes match at every input, but one graph is five units above the other.

Not every point on a graph has a finite derivative.

Mistake 1: Using the function value as the slope

For:

at \(x=3\) :

is the height. The tangent slope is:

Mistake 2: Writing a tangent equation without finding the point

A slope alone does not determine a unique line. Use both:

and:

Mistake 3: Assuming every horizontal tangent is a maximum or minimum

For:

the tangent at the origin is horizontal, but the function does not turn.

Mistake 4: Confusing a secant slope with a tangent slope

A secant uses two distinct curve points. A tangent slope comes from the derivative at one point.

Mistake 5: Using the sign of the function instead of the derivative

A graph can be below the \(x\) -axis and still have a positive slope. The sign of \(f(x)\) describes height. The sign of \(f'(x)\) describes tangent direction.

Mistake 6: Assuming a tangent cannot cross the curve

Tangency describes local slope agreement. Crossing is possible.

You should now be able to:

• Calculate a secant slope as an average rate of change.
• Interpret a tangent as the curve’s local direction.
• Explain that \(f'(a)\) is the tangent slope at \(x=a\).
• Distinguish the height \(f(a)\) from the slope \(f'(a)\).
• Find tangent slopes by differentiating and substituting.
• Write tangent equations using:

• Locate horizontal tangents by solving:

• Interpret positive, negative, and zero derivative values.
• Recognize that stationary points need not be turning points and that a derivative may fail to exist at a corner.