Derivative as an instantaneous rate of change

• Interpret a difference quotient as the slope of a secant line through two points.
• Explain how shrinking the horizontal interval produces a tangent-line slope.
• Evaluate the derivative of a quadratic at one point from the limit definition.

Build the idea

Worked example

1. Step 1

   Replace f of three plus h with the square of three plus h, and replace f of three with nine inside the difference quotient.

   $$\frac{(3+h)^2-9}{h}$$

2. Step 2

   Expand the squared binomial. The constant nine cancels with negative nine, leaving terms that both contain the nonzero interval width h.

   $$\frac{6h+h^2}{h}$$

3. Step 3

   For every nearby point with h not equal to zero, factor and cancel h. The secant slope now has the simpler form six plus h.

   $$\frac{h(6+h)}{h}=6+h$$

4. Step 4

   Let h approach zero in the simplified expression. The extra h contribution fades, and the secant slopes approach the tangent slope six.

   $$f'(3)=\lim_{h\to 0}(6+h)=6$$

• Substituting h equal to zero into the unsimplified difference quotient and treating zero divided by zero as a slope.

  The derivative is a limit of slopes for nonzero h values. Simplify the quotient on that punctured neighborhood first, then determine the value approached as h tends to zero.

• Expanding the square of three plus h as nine plus h squared and omitting the middle term.

  Use the full binomial identity: three plus h squared is nine plus six h plus h squared. The six h term is exactly what produces the nonzero tangent slope.

Use the same limit process to find the derivative of f of x equals x squared at x equals two.

This reviewed reading order covers the lesson explanation, equations, example, and checks without claiming that a video exists.

1. Part 1

   Start with a secant slope

   Fix a point on the graph at input a, then choose a second point h units away. The vertical change is f of a plus h minus f of a, and the horizontal change is h. Their quotient is the slope of the secant line and the average rate of change across that interval.

   • The quantity f of the quantity a plus h, minus f of a, end numerator, divided by h.
2. Part 2

   Move the points together

   Now let the second point slide toward the first. Each nonzero h creates a legitimate secant slope. As the gap shrinks, the secant lines rotate toward a tangent direction. The derivative exists when slopes from nearby inputs settle toward one value rather than jumping or growing without bound.

   • f prime of a equals the limit as h approaches zero of the quantity f of the quantity a plus h, minus f of a, end numerator, divided by h.
3. Part 3

   Expose the canceling interval

   For f of x equals x squared at three, substitute the two function values. Expanding gives the quantity six h plus h squared, divided by h. Factor h and cancel it while h is still nonzero. The slope between the two distinct points is now six plus h, which is defined for every nearby h.

   • The quantity three plus h, end quantity, squared, minus nine, all divided by h.
   • The quantity six h plus h squared, divided by h.
   • h times the quantity six plus h, divided by h, equals six plus h.
4. Part 4

   Read the limiting rate

   As h approaches zero, six plus h approaches six. This number is both the tangent-line slope and the instantaneous output change per input unit at x equal to three. Nearby numerical slopes such as six point one and six point zero one show the same convergence without ever setting the original denominator to zero.

   • f prime of three equals the limit as h approaches zero of six plus h, which equals six.

Scope and limitations

• This lesson considers a differentiable quadratic at one point and does not cover corners, cusps, vertical tangents, or discontinuities where a finite derivative may fail to exist.
• The visual uses positive intervals approaching from the right; a complete derivative requires compatible behavior from both sides when both sides are in the domain.